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Complex Numbers and Complex Functions
Mike PrestDepartment of Mathematics
University of ManchesterManchester M13 9PL
September 21, 2005
1
Contents
1 A bit of history 5
2 Complex numbers and their arithmetic 7
3 The complex plane 13
4 The unit circle and exponentiation 25
5 More on the arithmetic of complex numbers 29
6 Sine and cosine 34
7 Complex functions 39
8 The exponential and logarithm functions 46
9 Continuity and differentiability 52
10 Taylor and Laurent expansions: singularities and poles 61
11 Integration 66
12 Integration along parametrised contours 72
13 Theorems on integration 75
14 Real trigonometric integrals via complex numbers 83
15 Infinite real integrals 84
2
Introduction
These are course notes for the lecture course MT2101 Complex Numbers andComplex Functions. In the lectures I will explain what is in these notes and givemany more examples. The course itself is examples- rather than theory-basedin the sense that I will pay a lot more attention to computing examples thanto giving proofs of theorems. Nevertheless the theorems are necessary (theygive the tools for computing the examples) so it is important to understandwhy they are true (that is, to have some idea of their proofs) and to knowtheir limitations (so as not to try to apply them inappropriately since doing socan lead to false conclusions). If you want to see more of the, very beautiful,theory then you could try reading some of J. B. Conway, “Functions of OneComplex Variable”, Graduate Texts in Mathematics, Vol. 11, Springer-Verlag,1975 (which is in the university library and, despite being in the “GraduateTexts” series, is reasonably approachable, especially if you don’t worry too muchabout the details of the proofs). The book of Remmert mentioned below also isquite readable.
This course was given for many years by John B. Reade, and the book “Cal-culus with Complex Numbers”, Taylor and Francis, 2003 (ISBN 0-415-30847-X)is essentially his notes for the course. I have changed the course a bit, but notmuch, so that book (which is pretty inexpensive) is probably a worthwhile buy -it contains further examples, explains things somewhat differently and has someother supporting material.
These notes begin with what is essentially a review of the basics of complexnumbers (things you should know already) so in lectures I will actually movequite quickly through the earlier sections. In contrast, some of the later sectionslook rather short but I will actually spend quite a bit of time on them, usuallybecause I will be going through many examples. If time permits (and I hopeit does) I will add some extra material towards the end of the course (moreexamples and also more “theory”).
The exercise and solution sheets are in separate files. The first few exercisesheets contain much material which should be revision so use those exercisesaccordingly. You should attempt all exercises on new material. Solutions willbe given for most exercises.
I typed up these notes, exercises and solutions last year and the students,especially Sarah Jervis and Daniel Lopez, spotted quite a few typos and errors,which have now been corrected. But if you think you have found an error thatsurvived their scrutiny, do please let me know so that I can correct it. If youcome across any sources (websites, books) that you think might be useful toothers let me know, and I can add them to the website for this course.
These notes were prepared using the standard mathematical type-settingpackage TeX (the LaTeX dialect). Figures were prepared using XY-pic. Somefigures could be better but should serve their purpose. Some diagrams justwould have taken too long to prepare so a few are labelled TBCIC (To Be
3
Completed In Class) and there are also some places where I refer to othersources for illustrations.
4
1 A bit of history
Our idea of “number” has developed over thousands of years. The ordinarynumbers of counting, 1, 2, 3, . . . , are of ancient origin(s) and have been part ofall historical cultures. Other types of numbers gradually came to be used. Frac-tions, the “numbers” needed, for example, when sharing things out, were usedin ancient times. Dealing with profit and loss leads very naturally to negativenumbers but there was a long gap between their use in specific situations andthe acceptance of the general idea of “negative number”. Also the acceptanceof zero as a number took many years. In fact the picture is quite complicatedsince one can argue about what is meant by “acceptance” and this “acceptance”proceeded in different ways and at different rates in various cultures (and indeedin the minds of various individuals).
All the numbers so far mentioned have been rational numbers but the ancientGreeks also were aware of the need for irrational numbers, such as the squareroot of 2 and π (though no proof of the irrationality of π was known until thelater part of the 1700s).
Each stage of this process of extending the idea of “number” can be seen asarising from the need to solve specific types of equation (this is not to suggestthat it actually happened this way). For instance, to solve 3x = 5 we need tointroduce fractions, to solve x + 5 = 3 we need to introduce negative numbers,to solve x2 = 2 we need to introduce irrational numbers.
Complex numbers are necessary to solve x2+1 = 0. Unlike the other numbersmentioned they really were introduced in order to solve mathematical equations.Girolamo Cardano (1501-1576) posed the following problem. “Divide 10 intotwo parts, the product of which is 40.” You can check that the solutions to thisproblem are 5 ±
√−15. Cardano says: “... you will have to imagine
√−15,
... putting aside the mental tortures involved, a solution is obtained which istruly sophisticated”. More to the point, Cardano (and others, an interestingstory itself) found formulas for the solution of the general cubic equation but,when particular numbers are put in, the formula often results in expressionswhich are square roots of negative numbers, even when the cubic has three realroots. It was Rafael Bombelli (circa 1526-1573) who realised how to resolve thissituation. He treated (expressions for) complex numbers as valid and computedwith them using the ordinary rules of algebra. What he did seemed (and was)consistent and was the beginning of the acceptance of complex numbers as avalid part of mathematics.
You can read about this in various histories of mathematics (such as C.Boyer and U. C. Merzbach: A History of Mathematics, 2nd Ed., Wiley). A verythorough and readable account can be found in J.-P. Tignol, “Galois’ Theory ofAlgebraic Equations”, World Scientific (paperback edition 2001) (reading thiswill require a bit more commitment than reading the more general histories ofmathematics).
Complex numbers were increasingly used in mathematics in the 1600s and
5
1700s: Leonhard Euler (1707-1783) in particular made large advances. The ana-lytical aspects of complex numbers were hugely developed during the 1800s, es-pecially by (Baron) Augustin-Louis Cauchy (1789-1857), Georg Friedrich Bern-hard Riemann (1826-1866) and Karl Theodor Weierstrass (1815-1897). Againyou can get a quick account in general histories of mathematics and, if you wantto go deeper, there is a suitable book, Reinhold Remmert: Theory of ComplexFunctions, Graduate Texts in Mathematics, Vol. 122, Springer-Verlag, 1991.This is a (readable but hardly concise) textbook on complex analysis which hasa great deal of historical comment interspersed with the mathematics.
The usefulness of complex numbers in physics also came to be appreciated.One aspect was their usefulness in representing vectors in two dimensions. Var-ious people searched for a generalisation of complex numbers to some sort ofnumbers which could be used for representing vectors in three dimensions. (Sir)William Rowan Hamilton (1805-1865) found one solution (after eventually re-alising that the ideas he was developing needed a four-dimensional numbersystem). The system (the quaternions) that he came up with has i, j, k, allsquare roots of −1 but its multiplication is not commutative: ij = k, ji = −k,jk = i, kj = −i, ki = j, ik = −j. Hermann Gunther Grassmann (1809-1877)came to another solution, essentially the basis of multi-dimensional vector cal-culus.
A good website for the history of mathematics ishttp://www-gap.dcs.st- and.ac.uk/ history/
6
2 Complex numbers and their arithmetic
The equation x2 = −1 has no solution in the real numbers but it can be usefulto have a square root of −1 around, so we just introduce one, use i to denoteit and define a complex number to be an expression of the form a + bi wherea, b are real numbers.
The arithmetic of complex numbers is simply what we get by combining thearithmetic of real numbers with the equation i2 = −1.
•Addition: (a + bi) + (c + di) = (a + c) + (b + d)i
•Multiplication: (a + bi)(c + di) = (ac− bd) + (ad + bc)i
To check the formula for multiplication, multiply out the left hand side toget: ac + adi + bic + bidi = ac + adi + bci + bdi2 = ac + (ad + bc)i − bd =(ac− bd) + (ad + bc)i
The expressions a + bi, a + ib, ib + a, bi + a all denote the same complexnumber.
Note that −1 has two square roots, i and −i. There is complete symmetrybetween these: it makes no difference which square root we choose to denote byi (but, having made the choice, we should stick with it).
It is a remarkable fact that, by introducing a solution to the single equationx2 + 1 = 0, we have provided solutions to every quadratic equation. Namely,as you know, the generic quadratic equation ax2 + bx + c = 0 has its solutiongiven by the quadratic formula
x =−b±
√b2 − 4ac
2a.
Even if b2 − 4ac is negative it does have square roots (which are complexnumbers) since, if D > 0 then the square roots of −D are ±i
√D (check:
(±i√
D)2 = (±1)2i2(√
D)2 = −D).(By the way,
√D, where D > 0 is real, will, when it matters, denote the
positive square root of D.)Remember that if α1, α2 are the (possibly equal) values of
−b±√
b2 − 4ac
2a
the quadratic polynomial ax2 + bx + c factorises as (x − α1)(x − α2), so b =−α1 − α2 and c = α1α2.
Even more (than the fact that every quadratic with real coefficients hasa solution) is true: if the coefficients a, b, c of the quadratic polynomial areallowed to be complex then the quadratic formula still gives us solutions to the
7
equation ax2 + bx + c = 0. To show this, we have to check that every complexnumber has a square root which is also a complex number. (Actually, yet moreis true: every non-constant polynomial equation with complex coefficients hasa complex solution. This result is often called “the Fundamental Theorem ofAlgebra”. If we have some time we will show how this can be proved using theideas developed on the course.)
Example 2.1 Solve (1 + i)x2 + (2− 3i)x + (2− i) = 0 (1).We use the quadratic formula (which is valid for formal algebraic reasons
hence is valid for complex numbers as well as for real numbers).
x =−2 + 3i±
√(4− 9− 12i)− 4(2− i + 2i + 1)
2(1 + i)
=−2 + 3i±
√−17− 16i
2(1 + i)(∗).
So the problem reduces to finding the square root(s) of −17+16i. You mightalso wonder about the 1 + i on the bottom line. We’ll also deal with that issue.
How to find the square root of a complex number.We wish to calculate
√a + bi. Filled with optimism, we say: “Suppose that the
square root is, indeed, a complex number. Then it must have the form u + ivfor some real numbers u, v.” That is, u+ iv = ±
√a + bi. The problem is to find
the values of u and v in terms of the initial data a and b.Square both sides:
(u + iv)2 = (√
a + ib)2
that isu2 − v2 + 2uvi = a + bi.
Now, every equation between complex numbers is two equations betweenreal numbers:
a + bi = c + di iff{
a = c andb = d
so the above equation is equivalent to{u2 − v2 = a and2uv = b
Here a, b were given and u, v are unknowns whose values we’re trying to find.From the second equation we have v = b/(2u) (unless u = 0 but in that casethe equations reduce to −v2 = a, which we can solve directly).
Substituting into the first equation gives
u2 − b2
4u2= a
8
Multiply up and rearrange to obtain 4u4 − 4au2 − b2 = 0, which is a quadraticin u2 with solution
u2 =4a±
√16a2 + 16b2
8=
a±√
a2 + b2
2
Note that since a2 + b2 ≥ 0 the right-hand side of this equation is a realnumber.
Now we find u. It might seem, at first sight, that we have four possible valuesfor u but remember we want u (and v) to be real, so that cuts out one of theapparent possibilities, (a−
√a2 + b2)/2 (if different from (a+
√a2 + b2)/2, that
is, if b 6= 0), for u2.Therefore u2 = (a −
√a2 + b2)/2, hence we obtain two values (each the
negative of the other, possibly both zero) for u and, therefore, for v = b/(2u).
That’s how it works in principle. As for in practice:
Example 2.2 (Continued from Example 2.1) Find√−17− 16i. So, in the
notation of the general case just worked through, a = −17 and b = −16.Set u + iv =
√−17− 16i. Square both sides:
u2 − v2 + 2iuv = −17− 16iEquate real and imaginary parts:
u2 − v2 = −172uv = −16
By the second equation, u cannot be 0 so
v =−162u
=−8u
.
Substitute into the first equation to obtain
u2 − 64u2
= −17.
So u4 + 17u2 − 64 = 0 and hence
u2 =−17±
√172 + 2562
=−17±
√545
2.
Since u is to be real, u2 > 0 so discard −17−√
5452 to obtain u2 = −17+
√545
2(≈ 3.173).
Hence
u = ±
√−17 +
√545
2(≈ ±1.78)
and sov =
−8
±√
−17+√
5452
(≈ ∓4.49).
9
Therefore
√−17− 16i = ±
(√−17 +
√545
2− 8i√
−17+√
5452
)(≈ ±(1.78− 4.49i)).
Looking back at Example 2.2 we can substitute into equation (*) in Example2.1 and continue to compute the solutions to the quadratic equation (1). To dothis with the exact expressions above would now be a bit messy to write down.Also, how do we deal with the 1 + i on the bottom line? In other words, howcan we divide one complex number by another, writing (a + bi)/(c + di) in theform e + fi?
The “trick”, if you will, is to multiply top and bottom by the complexconjugate, c− di, of c + di:
a + bi
c + di=
a + bi
c + di· c− di
c− di=
(ac + bd) + (bc− ad)ic2 + d2
=ac + bd
c2 + d2+
bc− ad
c2 + d2i,
which does have the form e + fi for some real numbers e, f. (Note that thebottom line c2 +d2 is non-zero since c+di 6= 0 so at least one of c, d is non-zeroand hence c2 + d2 > 0.)
Example 2.3
2 + 3i
1− 12 i
=2 + 3i
1− 12 i·1 + 1
2 i
1 + 12 i
=(2− 3
2 ) + (3 + 22 )i
1 + ( 12 )2
=12 + 4i
54
=2 + 16i
5=
25
+165
i.
Example 2.4 We’ll finish off Example 2.1, 2.2 now but we will use approxima-tions since the expressions obtained in Example 2.2 for
√−17− 16i are rather
unwieldy and since the intention is to illustrate in very concrete terms that allquadratic polynomials with complex coefficients have complex roots.
In Example 1.2 we found√−17− 16i ≈ ±(1.78− 4.49i) (Note, by the way,
that, in contrast to square roots of positive real numbers, there is no sense inwhich one of these roots is “positive” and the other “negative”: all that we cansay is that each is −1 times the other.)
Substituting in (*) of Example 2.1 gives
x ≈ −2 + 3i± 1.78∓ 4.49i
2(1 + i)·(1− i
1− i
)=−2± 1.78 + (3∓ 4.49)i
2(1 + 1)· (1− i)
=(−0.22− 1.49i)(1− i)
4or
(−3.78 + 7.49i)(1− i)4
10
=(−0.22− 1.49)− (1.49− 0.22)i
4or
(−3.78 + 7.49) + (7.49 + 3.78)i4
=−1.71− 1.271i
4or
3.71 + 11.27i
4
≈ −0.43− 0.32i or 0.93 + 2.82i.
So the solutions of (1 + i)x2 + (2 − 3i)x + (−2 + i) = 0 are, to 1 decimalplace, −0.4− 0.3i and 0.9 + 2.8i
Summary:
i2 = −1
Typical complex number: z = a + bi (or a + ib, as you please)if b = 0 this is a real number
if a = 0 this is a (purely) imaginary number.
a is the real part of z = a + bi: Re(z) = ab is the imaginary part of z = a + bi: Im(z) = b
So z = Re(z) + iIm(z)
Equality of complex numbers: a + bi = c + di iff{
a = c andb = d
(one complex equation is equivalent to two real equations)
Addition: (a + bi) + (c + di) = (a + c) + (b + d)iMultiplication: (a + bi)(c + di) = (ac− bd) + (ad + bc)i
Division:
a + bi
c + di=
(a + bi)(c− di)(c + di)(c− di)
=ac + bd
c2 + d2+
bc− ad
c2 + d2i
(provided c + di 6= 0)
The (complex) conjugate of a + bi is a− bi.
More on conjugates
If z is a complex number we write z for its complex conjugate:
a + bi = a− bi.
Note that z + z is real (a + bi + a− bi = 2a = 2Re(z))and z − z is imaginary (a + bi− (a− bi) = 2bi = 2iIm(z))and z = z iff z is real (a + bi = a− bi iff b = 0)
11
Also z + w = z + wand zw = z w.
To prove these latter two, say z = a + bi and w = c + di. Then:z + w (add then conjugate) is
(a + bi) + (c + di) = (a + c) + (b + d)i = a + c− (b + d)i;also z + w (conjugate then add) is a + bi + c + di = a− bi + c− di which equalsthe expression for z + w.
The proof for multiplication is marginally less easy and should be done asan exercise.
Inverses
As usual, we write z−1 for 1/z (provided z 6= 0) so, if z = a + bi, then
z−1 =1
a + bi=
a− bi
(a + bi)(a− bi)=
a− bi
a2 + b2
or, noting that the top line is z,
z−1 =z
| z |2(∗∗)
where | z | is the modulus of z, defined by
|a + bi| =√
a2 + b2.
Multiply both sides of equation (**) by z to obtain
zz−1 =zz
|z|2
that is,
1 =zz
|z|2
and sozz = |z|2
an equation which can be useful.
Note that | − z| = |z|.Note also that
1i
= −i
(since 1i = 1
i ·ii = i
−1 )
A geometric interpretation of modulus is given later but note the followingkey property:
|zw| = |z||w|
12
To check this, let z = a + bi and w = c + di, so
|zw| = |(ac− bd) + (ad + bc)i| =√
(ac− bd)2 + (ad + bc)2
=√
a2c2 − 2acbd + b2d2 + a2d2 + 2adbc + b2c2 =√
a2c2 + b2d2 + a2d2 + b2c2.
On the other hand
|z||w| =√
a2 + b2√
c2 + d2 =√
(a2 + b2)(c2 + d2)
which, multiply it out, does equal the expression above.
3 The complex plane
A whole new dimension is added to our understanding of complex numbers byrepresenting them as points on a plane.
Each complex number a + bi is determined by its real and imaginary parts,a and b, so we can identify this number with the point (a, b) on a real plane.Taking this point of view, where we regard (a, b) ∈ R2 as a + ib ∈ C, we referto the complex plane.
//
OO
•i
•1
real axis
imaginary axis
The, as drawn here, horizontal axis consists of exactly the real numbersand is referred to as the real axis or real line. The vertical axis consists ofthe numbers of the form ib with b ∈ R so is called the imaginary axis orimaginary line.
Every point on this plane represents a complex number and vice versa sowe can think of this as a picture, usually called the Argand diagram, of thecomplex numbers.
13
//
OO
•i •1 + i
•1
•−1
• −i
Addition of complex numbers can be understood as addition of vectors inthe complex plane
//
OO
◦◦
44444444444
ZZ•−2 + 3i ������������
KK•1 + 4i
jjjjjjjjjj
44• 3 + i
jjjjj
44
44
44
and complex conjugation is reflection in the real axis
//
OO
ooooooooooooo
77• 2 + i
OOOOOOO
''• 2− i
This point of view makes the triangle inequality |z+w| ≤ |z|+ |w| obvious
//
OO�������������������
◦z
|z|
;;;;;;;;;;;;;;
◦w
mmmmmmmmm
◦z + w|z + w|
;;
;;
;;
;
|w|
��
��
��
��
��
14
since |z + w| is the length of one side of the upper triangle, which is less thanthe sum of the lengths of the other two sides.(This can, of course, be proved algebraically, by writing z as a + bi and w asc+di, but the algebraic proof, which requires rather more work than one mightexpect, is left as an exercise.)
There are useful variations on this inequality.
First, in |z + w| ≤ |z|+ |w| replace z by z − w to get
|z − w + w| ≤ |z − w|+ |w|
which rearranges to|z| − |w| ≤ |z − w| (1).
Switch the roles of z and w to obtain
|w| − |z| ≤ |w − z|.
But |w − z| = | − (z − w)| = |z − w| so also
|w| − |z| ≤ |z − w| (1).
Combine equations (1) and (2) (note that at most one of |z| − |w|, |w| − |z| isstrictly positive) to obtain ∣∣|z| − |w|∣∣ ≤ |z − w|.
Now replace w by −w in (1) and, noting that | − w| = |w|, deduce
|z| − |w| ≤ |z + w|.
Example 3.1 Show that for all z with |z| = 4 we have
723
≤∣∣∣ 2z − 1z2 + z − 3
∣∣∣ ≤ 1.
Consider the top line first:
|2z − 1| ≤ |2z|+ 1 = 2|z|+ |1| = 8 + 1 = 9.
Also |2z − 1| ≥ |2z| − |1| = 8− 1 = 7,so
7 ≤ |2z − 1| ≤ 9.
As for the bottom line,
|z2 + z − 3| ≤ |z2 + z|+ |3| ≤ |z2|+ |z|+ 3 = |z|2 + |z|+ 3 = 16 + 4 + 3 = 23
15
and|z2 + z − 3| ≥ |z2 + z| − |3| ≥ |z2| − |z| − 3 = 16− 4− 3 = 9
so9 ≤ |z2 + z − 3| ≤ 23.
Therefore19≥∣∣∣ 1z2 + z − 3
∣∣∣ ≥ 123
and so723
≤ |2z − 1||z2 + z − 3|
=∣∣∣ 2z − 1z2 + z − 3
∣∣∣ ≤ 99
= 1.
Example 3.2 Show that the complex numbers z which satisfy the equation∣∣∣ z − i
2z + 1
∣∣∣ = 2√
2√7
form a circle, and find its centre and radius.
First set z = x + iy, so we have
2√
2√7
=∣∣∣ x + iy − i
2x + 2iy + 1
∣∣∣ = |x + i(y − 1)||(2x + 1) + 2iy|
.
Square both sides and multiply up to obtain8|(2x + 1) + 2iy|2 = 7|x + i(y − 1)|2,
that is,8((2x + 1)2 + (2y)2) = 7(x2 + (y − 1)2),
that is,8(4x2 + 4x + 1 + 4y2) = 7(x2 + y2 − 2y + 1),
that is25x2 + 32x + 25y2 + 14y = −1.
Complete the squares to obtain(5x +
165
)2
− 25625
+(5y +
75
)2
− 4925
= −1
that is (5x +
165
)2
+(5y +
75
)2
=256 + 49− 25
25=
28025
.
Now, the square of the distance of a complex number x + iy from a complexnumber a+ ib is, by Pythagoras’ Theorem, (x−a)2 +(y− b)2 (see the diagram).
16
//
OO
�����������������•
•
a + ib
x + iy
Therefore the equation satisfied by those complex numbers z = x + iy at adistance s from a + ib (i.e. the equation of the circle in the complex plane withcentre a + ib and radius s) is
(x− a)2 + (y − b)2 = s2.
Clearly we can bring our equation in this form if we first divide it by 52 = 25to obtain (
x +1625
)2
+(y +
725
)2
=(√280
25
)2
and then rearrange to obtain the form where we can just read off the centre andradius: (
x−(− 16
25))2
+(y −
(− 7
25))2
=(2√
7025
)2
.
This is the equation of the circle with centre − 1625 −
725 i and radius 2
√70
25 .
Multiplication is better understood/pictured using polar coordinates, whichwe discuss now.
One polar coordinate is the distance, r, of a point from the origin. In thecontext of complex numbers this is the modulus, r =
√a2 + b2 = |z| of the
complex number z = a + ib: see the diagram.
//
OO
•a
•ib uuuuuuuuuuuuuu
• a + ib
r
so r =√
a2 + b2 = |a + ib|. For example:
17
//
OO
◦
◦
*******************
•−1 + 3i
ooooooooooooo
•−2− i
| − 1 + 3i| =√
1 + 9 =√
10| − 2− i| =
√4 + 1 =
√5
The other polar coordinate is the angle measured with respect to a fixedreference half-line (that is, the “bearing” of a point, as measured from theorigin, from a fixed reference direction). The fixed line will always be takento be the positive real half-line and our convention will be to measure anti-clockwise rotation. This second coordinate, the angle (measured in radiansunless measurement in degrees is specified), is typically denoted θ and is calledthe argument of z, denoted arg(z).
//
1111111
XX
0 1◦ ◦
z
θ=arg(z)ss
For example:
//
OO
���������
◦1 + i
π4
ee//
OO
?????????
◦−1 + i3π4rr
//
OO
���������
◦−1− i
5π4
3π4
jj //
OO
?????????
◦1− i
7π4
55 −π4xx
18
An important point, illustrated in the last example above, is that arg(z) isnot single-valued, because a complete rotation, in either the anticlockwise (posi-tive) or clockwise (negative) sense leaves all bearings unchanged so, if arg(z) = θthen, for every integer k, also arg(z) = θ+2kπ (since 2π represents one completerotation). Therefore the symbol “=” in an expression such as “arg(z) = θ” isweaker than true equality and should, strictly speaking, be read as “one valueof arg(z) is θ”.
The value of arg(z) lying in the interval [0, 2π) is what we will call theprincipal (or standard) value of arg(z) (some people prefer to choose theprincipal value to lie in the interval (−π, π]). For instance, the principal valueof arg(1− i) is, by this convention, 7π
4 (not −π4 or 15π
4 or . . . ).Therefore, writing
arg(1 + i) =π
4+ 2kπ (k ∈ Z)
gives the full story.
Notice the important special cases
arg(0) = (0+)2kπ; arg(i) =π
2+ 2kπ;
arg(−1) = π + 2kπ,= (2k + 1)π; arg(−i) =3π
2+ 2kπ,= −π
2+ 2lπ
where, throughout, k, l ∈ Z
//
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◦iπ2
ii
π2 +2π
mm
The multivalued nature of arg(z) is of much greater significance than onemight at first expect. For example see where we compute n-th roots of complexnumbers and also where we discuss the exponential and logarithm functions.
Notice (draw the diagram) that if |z| = r and arg(z) = θ then
z = rcos(θ) + irsin(θ).
Recall that if we multiply two complex numbers together then the modulusof the product is the product of their moduli:
|zw| = |z| |w|.
19
Also, the argument of the product is the sum of their arguments:
arg(zw) = arg(z) + arg(w).
To prove the latter, say z = rcos(θ)+irsin(θ) and w = scos(φ)+is sin(φ). Then
zw = rcos(θ)scos(φ)− rsin(θ)s sin(φ) + i(rcos(θ)s sin(φ) + rsin(θ)scos(φ)
= rscos(θ + φ) + irs sin(θ + φ)
(using the trigonometric formulas for the sine and cosine of a sum of angles).
Since zz−1 = 1 has argument 0 it follows that
arg(z−1) = −arg(z).
Also note thatarg(z) = −arg(z).
Moving between cartesian and polar coordinates
Given a complex number z = a + ib, the modulus of z is given by |z| =√a2 + b2 and, if z 6= 0, the argument of z is arg(z) = θ + 2kπ where θ is any
angle satisfying
sin(θ) =b√
a2 + b2and cos(θ) =
a√a2 + b2
(∗ ∗ ∗)
(there will be just one such θ in the interval [0, 2π)).
This is obvious if z lies in the first quadrant
//
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◦a
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◦ a + ib
θee
but, otherwise, requires you to recall how the signs of cos(θ) and sin(θ) vary inthe interval [0, 2π)
//
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◦ ib
◦a
8888888888888
◦a + ib
θpp
20
In the example above a is negative so cos(θ) will be negative.
Writing θ = tan−1(
ba
)is not enough (though it is true that tan(θ) = b
a )because there are two values of θ in [0, 2π) satisfying this equation.
Example 3.3 Take z = −1− i so tan(arg(z)) = −1−1 = 1. This does not deter-
mine arg(z), which is 5π4 , because also tanπ
4 = 1.
//
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y = tan(x)
π2
/π
•5π4
3π2
/
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In other words, if z = a + ib then to determine θ = arg(z) we can computeba , hence tan−1( b
a ) but we also have to look at the sign of cos(θ) (i.e. a√a2+b2
,
hence the sign of a) or the sign of sin(θ) (i.e. b√a2+b2
, hence the sign of b) topin down θ within the interval [0, 2π).
Example 3.4 Continuing with the above example, with z = −1 − i, we havetan(arg(z)) = 1. Now note that the sign of cos(arg(z)) is equal to the sign of thereal part of z, which is negative. Thinking of the graph of the cosine function,we see that θ ∈ (π
2 , 3π2 )
21
//
OO
◦π2 ◦
π◦3π3
y = cos(x)
so, combined with the information that tan(θ) = 1, hence θ = π4 or 5π
4 , thisgives θ = 5π
4 .
It can be useful to describe θ = arg(z) in the form: angle in [0, π2 )+ mul-
tiple of π2 . This is just basic trigonometry but we remind you that it is easy
if you remember the shapes and symmetries of the graphs of the trigonometricfunctions.
//x◦
OOy
//x
//x
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−π2
π2
π3π2
2π
y = sin(x)
y = cos(x)
y = tan(x)
For t ≥ 0 we will let tan−1(t) denote the angle between 0 and π2 with tangent
t.
22
Example 3.5 Find the moduli and arguments of the following complex num-bers.
(i) −2 + 13 i
(ii) −2− i2√
3(iii) −5i
Answers
(i) | − 2 + 13 i| =
√4 + 1
9 =√
379 =
√37/3.
Let θ1 = arg(−2 + 13 i) so
tan(θ1) =1/3−2
= −16.
Now, sin(θ1) is positive (it has the same sign as 13) and cos(θ1) is negative
(same sign as −2), which puts θ1 between π2 and π.
Also, tan(θ1) = − 16 = −tan−1( 1
6 ) (where tan−1( 16 ) is, by our convention, in
[0, π2 )) so, by symmetry of the graph of y = tan(x), we have θ1 = π− tan−1( 1
6 ).Hence
arg(−2+13i) = π−tan−1(
16)+2kπ (k ∈ Z) (= −tan−1(
16)+(2k+1)π (k ∈ Z)).
(ii) | − 2− i2√
3| =√
4 + 4 · 3 =√
16 = 4.If θ2 = arg(−2− i2
√3) then
tan(θ2) =−2√
3−2
=√
3.
Both sin(θ2) and cos(θ2) are negative, which means that (the principal valueof) θ2 is in the interval (π, 3π
2 ] and tan(θ2) =√
3 = tan(π3 ) (see the triangle
below). So, by symmetries of the graph of y = tan(x), θ2 = π + tan−1√
3 =π + π
3 = 4π3 .
Therefore
arg(−2− i2√
3) =4π
3+ 2kπ (k ∈ Z) (=
π
3+ (2k + 1)π (k ∈ Z)).
(iii) | − 5i| = | − 5||i| = 5 · 1 = 5,Let θ3 = arg(−5i): then tan(θ3) is undefined (“−5
0 ”). The value of sin(θ3) isnegative and cos(θ3) = 0 so, from the graphs of these trigonometric functions,we see that (the principal value of) θ3 is 3π
2 .Hence
arg(−5i) =3π
2+ 2kπ (k ∈ Z) (=
π
2+ (2k + 1)π (k ∈ Z)).
23
The triangle shown is useful for working out the values of the trigonometricfunctions at π/3 and π/6.
______
22
22
22
22
22
22
1
2√
3
π3
aa
Moving from polar to cartesian form is easier than vice versa since, if |z| = rand (one value of) arg(z) = θ, then, as noted earlier, we have
a = rcos(θ) and b = rsin(θ).
That is, if |z| = r and arg(z) = θ then
z = rcos(θ) + irsin(θ) (∗ ∗ ∗∗)
Example 3.6 Find z in cartesian form in the following cases.(i) |z| = 4, arg(z) = π
3(ii) |z| = 1√
2, arg(z) = −π
4
(iii) |z| = 3.4, arg(z) = 2
Answers
(i) cos(arg(z)) = cos(π3 ) = 1
2 and sin(arg(z)) =√
32 (see the triangle above)
so
z = 4× 12
+ i4√
32
= 2 + i2√
3.
(ii) cos(−π4 ) = cos(π
4 ) = 1√2
and sin(−π4 ) = −sin(π
4 ) = − 1√2
(note, by theway, that the value of arg(z) given was not the principle value since −π
4 /∈[0, 2π)) so
z =1√2· 1√
2+ i
1√2· −1√
2=
12− i
2.
(iii) (Note that arg(z) = 2 lies between π2 (≈ 1.57) and π so arg(z), being in
the second quadrant, will have positive sine and negative cosine.)We have
z = 3.4cos(2) + i3.4sin(2).
Putting in approximate values for cos(2) and sin(2) gives the approximation
z ≈ 3.4× (−0.42) + i3.4× (0.91) ≈ −1.4 + 3.1i.
24
Summary
z = a + ib = rcos(θ) + irsin(θ) where
r = |z| =√
a2 + b2 and θ = arg(z) satisfies
{cos(θ) = a√
a2+b2
sin(θ) = b√a2+b2
A special case is z = 0, where |z| = 0 but arg(z) is undefined.
It is interesting to look at the lines and curves where one of these coordinatefunctions is constant.
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a/
Re(z) = a
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__________b
/Im(z) = b
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◦r
◦ ir
◦−r
◦−ir
| z |= r
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arg(z) = θθuu
Note in particular that points with fixed modulus r lie on a circle, centredat the origin, of radius r and the points with fixed argument lie on a half linegoing from, but not including, the origin.
The modulus of a complex number is its “size”. If we know |z| then we knowthe circle (centre 0) that it lies on. If we also know the angle arg(z) then wecan fix the point z exactly (it is the intersection of the circle |z| = r with theline arg(z) = θ).
In the next section we will give particular prominence to the unit circle(that is, the circle, C(0, 1), with centre 0 and radius 1).
4 The unit circle and exponentiation
Recall that if t is a real number then
et = 1 + t +t2
2!+
t3
3!+ · · ·+ tn
n!+ · · · =
∞∑n=0
tn
n!
25
(where at n = 0 we use the convention that 0! = 1 so t0/0! = 1)and this series is convergent for all values of t. By this we mean that the partialsums
SN =N∑
n=0
tn
n!
tend to a limit as N →∞, so they approximate et in the sense that the differ-ences
et − SN =∞∑
n=N+1
tn
n!
go to 0 as N →∞.
Suppose we replace the real number t by the purely imaginary number itand so define
eit = 1 + (it) +(it)2
2!+
(it)3
3!+ · · ·+ (it)n
n!+ . . .
= 1 + it− t2
2!− it3
3!+
t4
4!+
it5
5!− t6
6!− . . . .
Gathering together real and imaginary parts (one can prove that this re-arrangement is valid)
eit =(1− t2
2!+
t4
4!− t6
6!+ . . .
)+ i(t− t3
3!+
t5
5!+ . . .
)= cos(t) + isin(t),
where we recall that
cos(t) = 1− t2
2!+
t4
4!− t6
6!+ · · · =
∞∑n=0
(−1)n t2n
(2n)!
sin(t) = t− t3
3!+
t5
5!− · · · =
∞∑n=0
(−1)n t2n+1
(2n + 1)!
The resulting formula
eit = cos(t) + isin(t) t ∈ R
is a remarkable one, which we will use time and again.
First, it gives another way of writing any complex number: the earlier for-mula (****) z = rcos(θ)+ irsin(θ), that is, z = r(cos(θ)+ isin(θ)) where r = |z|and θ = arg(z), becomes
z = reiθ,
that isz = |z|eiarg(z).
26
This is also referred to as the polar form of a complex number.
It is the case, although we will not prove it here, that the usual rules for powerseries are valid (roughly, that if a series is absolutely convergent(= convergentin modulus) then we can do whatever we like to it), and we will use these freely.
Example 4.1 Referring back to Example 3.5 we have(i)
−2 +13i =
√373
ei(π−tan−1( 16 ))
We can rewrite the RHS as√
373
eiπe−itan−1( 16 ) = −
√373
e−itan−1( 16 )
since eiπ = cos(π) + isin(π) = −1.
(ii)−4− 2
√3i = 2
√7e
4πi3
(iii)−5i = 5e
3πi2
There are a number of points to observe here.
First, arg(z) is defined only up to addition of multiples of 2π, so what effectdoes this have on eiarg(z)? Well, e2πi = cos(2π) + isin(2π) = 1. (Notice howmany of the fundamental constants of mathematics make an appearance in thisformula e2πi − 1 = 0.)
Soei(θ+2kπ) = eiθe2kπi = eiθ(e2πi)k = eiθ1k = eiθ
and, therefore, any choice of arg(z) gives the same value for eiarg(z) as theprincipal value.
Next, some notable special cases.
e2kπi = 1, in particular e2πi = 1 = e0
eπ2 i = i eπi = −1 e
3π2 i = e−
π2 i = −i
All these values lie on the unit circle. In fact,
|eiθ| = |cos(θ) + isin(θ)| =√
cos2(θ) + sin2(θ) = 1
and, conversely, any complex number on the unit circle has modulus 1 so hasthe form
cos(θ) + isin(θ) = eiθ.
27
//
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◦ i
◦−1
◦−i
◦�����
___ z = cos(θ) + isin(θ)= reiθ
◦isin(θ)
◦cos(θ)
θjj
That is:
the unit circle in the complex plane consists exactly
of the complex numbers of the form eiθ.
Notice that the mapping which takes the real parameter θ to the complexnumber eiθ is periodic, of period 2π. Indeed, it wraps the real line around thecomplex unit circle infinitely many times (with, under our conventions, thepositive θ direction corresponding to winding in the anticlockwise direction).
The representation z = reiθ can, therefore, be thought of as: starting at theorigin, find the point, eiθ, on the unit circle in the same direction as z, thenscale by r = |z| to obtain z.
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28
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Summary
eit = cos(t) + isin(t)
z = |z|eiarg(z)
5 More on the arithmetic of complex numbers
The geometric interpretation of addition of complex numbers has been discussedalready. Multiplication is best understood in terms of polar coordinates.
First consider the effect of multiplication by i. The typical complex numberhas the form z = reiθ. So
zi = reiθ · ei π2 = rei(θ+ π
2 ).
This complex number has the same modulus as z but its argument has beenincreased by π
2 . In other words, the effect of multiplication by i is rotation an-ticlockwise by π
2 .
//
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rrrrrrr
◦ z2222222
◦iz
uu 22rr
29
More generally, if z = reiθ and w = seiφ then zw = reiθseiφ = rsei(θ+φ).So, as we have seen before,
|zw| = |z||w| and arg(zw) = arg(z) + arg(w) (∗5)
Therefore, fixing z = reiθ, the effect on complex numbers of multiplying byz is to rotate anticlockwise through an angle θ and scale by a factor of r.
Example 5.1 Compute (1 + i)100
Although it is possible to expand this using the binomial theorem and thensimplify, this method is not recommended! Much easier is first to convert 1 + ito polar form:
|1 + i| =√
1 + 1 =√
2
arg(1 + i) =π
4So
1 + i =√
2ei π4 .
Then (1 + i)100 = (√
2ei π4 )100 =
√2100
(ei π4 )100 = 250e25πi = 250e24πieπi =
250eπi = −250 (since eπi = −1).
We also have the following (which we discussed earlier, using different nota-tion):
if z = reiθ then z = re−iθ and z−1 = r−1e−iθ.
To check these: we already know zz = |z|2 = r2 so
z =r2
reiθ(assuming z 6= 0) =
r
eiθ= re−iθ
andz−1 = 1/(reiθ) = r−1e−iθ.
Example 5.2 Compute (1 + i)−1
We have (see above) 1 + i =√
2ei π4 so
(1 + i)−1 =1√2e−i π
4
(= 1√2ei 7π
4 to use the principal value of the argument)
=1√2(cos(−π
4) + isin(−π
4))
=1√2(cos(
π
4)− isin(
π
4)) =
1√2(
1√2− 1√
2i) =
12− 1
2i.
An alternative computation is:
11 + i
=1− i
(1 + i)(1− i)=
1− i
1 + 1=
12− 1
2i.
30
Example 5.3
5ei 3π2 · 1√
2ei 3π
4 =5√2ei( 3π
2 + 3π4 ) =
5√2ei( 9π
4 ) =5√2ei(2π+ π
4 ) =5√2e2πiei π
4
=5√2(1 · ei π
4 ) =5√2ei π
4
(=
5√2(
1√2
+ i1√2) =
52
+ i52
).
De Moivre’s Theorem is the equality
(cos(θ) + isin(θ))n = cos(nθ) + isin(nθ)
To prove this: LHS =(eiθ)n = einθ = RHS
This is very useful for generating trigonometric identities.
Example 5.4 n = 2cos(2θ) + isin(2θ) = (cos(θ) + isin(θ))2
= cos2(θ) + 2icos(θ)sin(θ) + (isin(θ))2
= cos2(θ)− sin2(θ) + 2icos(θ)sin(θ).Now equate real and imaginary parts to conclude
cos(2θ) = cos2(θ)− sin2(θ) andsin(2θ) = 2cos(θ)sin(θ)
Example 5.5 n = 3cos(3θ) + isin(3θ) = (cos(θ) + isin(θ))3
= cos3(θ) + 3cos2(θ)isin(θ) + 3cos(θ)(isin(θ))2 + (isin(θ))3
= cos3(θ) + 3icos2(θ)sin(θ)− 3cos(θ)sin2(θ)− isin3(θ)Equating real and imaginary parts gives
cos(3θ) = cos3(θ)− 3cos(θ)sin2(θ)= cos3(θ)− 3cos(θ)(1− cos2(θ)) = 4cos3(θ)− 3cos(θ) andsin(3θ) = 3cos2(θ)sin(θ)− sin3(θ)= 3(1− sin2(θ))sin(θ)− sin3(θ) = 3sin(θ)− 4sin3(θ).
Similarly, for every positive integer n, cos(nθ) and sin(nθ) may be written interms of cos(θ) and sin(θ). (Such formulas were derived, using another method,by Francoise Viete (1540-1603).)
Polar form also may be used to generate other trigonometric formulas.
Example 5.6 cos(A + B) + isin(A + B) = ei(A+B)
= eiAeiB = (cos(A) + isin(A))(cos(B) + isin(B))= cos(A)cos(B) + isin(A)cos(B) + icos(A)sin(B)− sin(A)sin(B).
Equating real and imaginary parts givescos(A + B) = cos(A)cos(B)− sin(A)sin(B) andsin(A + B) = sin(A)cos(B) + cos(A)sin(B).
31
Taking n-th roots of complex numbers is straightforward using polar form(it also shows the importance of the fact that arg(z) is multivalued).
Example 5.7 Find all cube roots of 1.
We have 1 = ei0 so if z is a cube root of 1 then z3 = 1 = ei0. In particular|z3| = 1 so |z|3 = 1 and hence |z| = 1
13 = 1. Since, moreover
arg(z3) = 3arg(z)
(by (*5)) we have
arg(z) =13arg(z3) =
13arg(1).
Now remember thatarg(1) = 0 + 2kπ (k ∈ Z)
soarg(z) =
13(0 + 2kπ) =
2kπ
3(k ∈ Z)
= . . . 0,2π
3,4π
3,6π
3
(= 2π + 0
),8π
3
(= 2π +
2π
3
),10π
3
(= 2π +
4π
3
), . . .
You can see that, up to addition of multiples of 2π, there are three possiblevalues of arg(z), namely 0, 2π
3 , 4π3 .
Therefore the (three) cube roots of 1 are:
1ei0, 1ei 2π3 , 1ei 4π
3
that is, 1, e2πi3 , e
4πi3 (= (e
2πi3 )2 = e−
2πi3 ).
Plotting these on the Argand diagram, we see that they are equally spacedround the unit circle.
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11
1
•ei 2π3
•ei 4π
3
2π3vv
Notice that the sum of the three different cube roots of 1 is 0:
1 + e2πi3 + e
4πi3 = 1 +
(cos(2π
3
)+ isin
(2π
3
))+(
cos(4π
3
)+ isin
(4π
3
))
= 1− 12
+ i
√3
2− 1
2− i
√3
2= 0.
32
But there’s a simpler way to see this, as follows.Let ω be either of the non-real cube roots of 1. Then (ω2)3 = ω6 = (ω3)2 =
12 = 1 so w2 also is a cube root of 1 (and cannot equal 1 since (ω2)2 = ω4 = ω)so, therefore, ω, ω2, 1 are the three distinct cube roots of 1. Then ω(ω+ω2+1) =ω2 + ω3 + ω = ω2 + 1 + ω = ω + ω2 + 1 so, since ω 6= 1, it must be that
ω + ω2 + 1 = 0.
Example 5.8 Find all the cube roots of −1.
We have −1 = eiπ.As in Example 5.7 if z3 = −1 then |z| = | − 1| 13 = 1 and
arg(z) =13(π + 2kπ) = . . . ,
π
3,3π
3,5π
3,7π
3(= 2π +
π
3), . . .
giving, modulo integer multiples of 2π, exactly three values π3 , π, 5π
3 .
So the cube roots of −1 are:
ei π3 , eiπ (= −1), ei 5π
3 .
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•
11
1
•
Example 5.9 Find all 5th roots of 1− i
//
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??
?
◦
First, write 1− i in polar form:1− i =
√2ei(−π
4 ) (=√
2ei 7π4 if you prefer).
So
(1− i)15 =
√2
15 (ei(−π
4 +2kπ))15 (k ∈ Z) = (2
12 )
15 ei(− π
20+ 2kπ5 ).
33
There will be five distinct fifth roots of 1−i so substitute five successive valuesof k to get these, say k = 0, 1, . . . , 4:
2110 e−
π20 i, 2
110 e(− π
20+ 2π5 )i, 2
110 e(− π
20+ 4π5 )i, 2
110 e(− π
20+ 6π5 )i, 2
110 e(− π
20+ 8π5 )i
that is, 2110 e−
πi20 , 2
110 e
7πi20 , 2
110 e
15πi20 , 2
110 e
23πi20 , 2
110 e
31πi20 .
These are indicated on the following diagram.
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ET_
•
••
•• ◦ 1− i
Summary
|zw| = |z||w| and arg(zw) = arg(z) + arg(w)
(cos(θ) + isin(θ))n = cos(nθ) + isin(nθ) (De Moivre)
if z = reiθ then z1n = r
1n ei(
θn + 2kπ
n
)(k = 0, 1, . . . , n− 1)
6 Sine and cosine
From the fundamental formula
eit = cos(t) + isin(t)
we have, replacing t by −t,
e−it = cos(−t) + isin(−t) = cos(t)− isin(t).
Solving these two equations for sin(t) and cos(t) we obtain Euler’s Formulas:
cos(t) =12(eit + e−it)
sin(t) =12i
(eit − e−it).
We used De Moivre’s Theorem to express cos(nθ) and sin(nθ) in terms ofcos(θ) and sin(θ). Euler’s formula can be used for the reverse direction.
34
Example 6.1 Write cos5(t) in terms of cos(nt) for various values of n.
cos5(t) =(
12 (eit + e−it)
)5
=125
((eit)5+5(eit)4(e−it)+10(eit)3(e−it)2+10(eit)2(e−it)3+5(eit)(e−it)4+(e−it)5
)=
125
(e5it + 5e3it + 10eit + 10e−it + 5e−3it + e−5it
).
Now comes the clever bit:
=125
(ei5t + e−i5t + 5(ei3t + e−i3t) + 10(eit + e−it)
)=
125
(2cos(5t) + 5.2cos(3t) + 10.2cos(t)
)(by Euler’s Formula)
=124
cos(5t) +524
cos(3t) +523
cos(t).
Example 6.2 Find I =∫ π
20
cos5(t)dt
By the above example
I =124
∫ π2
0
(cos(5t) + 5cos(3t) + 10cos(t)
)dt
=124
[15sin(5t) +
53sin(3t) + 10sin(t)
]π2
0
=124
[15sin(5π
2
)+
53sin(3π
2
)+ 10sin
(π
2
)− 0]
=124
(15− 5
3+ 10
)=
124
(3− 25 + 150
15
)=
12816 · 15
=815
.
Example 6.3 Find I =∫ π
0cos7(t)dt.
Think first! and remember the symmetries of the graph of cos(t).
I =∫ π
2
0
cos7(t)dt +∫ π
π2
cos7(t)dt
=∫ π
2
0
cos7(t)dt +∫ π
2
0
cos7(t +π
2)dt =
∫ π2
0
cos7(t) + cos7(t +π
2).
But cos(t + π2 ) = −cos(t) so cos7(t + π
2 ) = −cos7(t) and hence
I =∫ π
2
0
0dt = 0.
(That is, the positive and negative areas under the graph cancel out.)
35
In Euler’s Formulas t stands for a real number but it also makes sense toallow a complex number in place of t.
First we note that ez makes sense when z is complex, say z = a + ib. Forwe have ez = ea+ib = eaeib. The term ea is a real number (since a is real) andeib = cos(b) + isin(b) has already been defined. Therefore
ea+ib = eacos(b) + ieasin(b).
So, for instance, eiz = ei(a+ib) = eia−b = e−beia.Therefore we just define
cos(z) =12(eiz + e−iz)
sin(z) =12i
(eiz − e−iz).
It is still the case that the expected power series expansions for these func-tions are valid: just as
ez =∞∑
n=0
zn
n!
so also
cos(z) =∞∑
n=0
(−1)n z2n
(2n)!
and
sin(z) =∞∑
n=0
(−1)n z2n+1
(2n + 1)!.
You can check easily enough that these equalities are formally valid butthere is an issue of convergence, and even of what is meant by an infinitesum of complex numbers. In fact, just as in the real case, if z0, z1, . . . , zn, . . .are complex numbers then we define
∑∞n=0 zn to be the limit of partial sums
limN→∞∑N
n=0 zn if this exists (if it doesn’t we just say that∑∞
n=0 zn is un-defined). That pushes back the problem to what we mean when we say thatthe limit of a sequence, (SN =
∑Nn=0 zn)N→∞, of complex numbers exists. But
again the solution is just as in the reals: such a sequence will be convergent iffit is a Cauchy sequence. (Precisely, one requires that for every ε > 0 there is N
large enough so that all the subsequent differences sN+K − sN =∑N+K
n=N+1 zn
are less than ε in modulus, that is, written formally,
∀ε > 0∃N ∀K |SN+K − SN | < ε.
The only difference with the real case is that here modulus rather than absolutevalue is used.)
36
Example 6.4 cos(i) = 12 (ei2 + e−i2) = 1
2 (e−1 + e1) ≈ 1.54sin(i) = 1
2i (ei2 − e−i2) = −i
2 (e−1 − e1) ≈ 1.18i
The values of cosine and sine for real arguments lie between −1 and +1(hence between 0 and 1 in modulus), but this is no longer true for complexarguments. In fact, the moduli of cos(z) and sin(z) as z varies in C are un-bounded.
For instance,
cos(iy) =12(e−y + ey) →∞ as y →∞.
You might notice that the right-hand side of the above equation is cosh(y)(= 1
2 (ey + e−y)). Just like the trigonometric functions, the hyperbolic functionsmay be extended to allow complex arguments. We set
cosh(z) =12(ez + e−z)
sinh(z) =12(ez − e−z).
(Just as with real arguments these can be expressed as power series:
cosh(z) =∞∑
n=0
z2n
(2n)!
and
sinh(z) =∞∑
n=0
z2n+1
(2n + 1)!.)
Then the following identities are easily checked:
cos(iz) = cosh(z) cosh(iz) = cos(z)
sin(iz) = isinh(z) sinh(iz) = isin(z).
Also, if z = x + iy, then, using that the standard trigonometric identitiesare also valid for complex arguments (they are formal consequences of the defi-nitions),
cos(z) = cos(x + iy) = cos(x)cos(iy)− sin(x)sin(iy)
= cos(x)cosh(y)− isin(x)sinh(y)
and similarly (exercise) for sin(z).
37
Example 6.5 Find the (complex) values of z for which cos(z) = 0.
Set z = x + iy so (see above) cos(z) = cos(x)cosh(y)− isin(x)sinh(y).Therefore, cos(z) = 0 iff cos(x)cosh(y) = 0 = sin(x)sinh(y)iff cos(x) = 0 (remember |cosh(y)| ≥ 1) and sin(x)sinh(y) = 0.
Now, cos(x) = 0 iff x = π2 + nπ for some n ∈ Z. Also, where cos(x) = 0 we
have sin(x) 6= 0 (because cos2 + sin2 = 1) so for such values sin(x)sinh(y) = 0iff sinh(y) = 0 and this happens iff y = 0 (recall the graph of sinh(y)).
Therefore the complex values of z with cos(z) = 0 are z = π2 + nπ (n ∈ Z)
(so, in fact, are just the real values of z with cos(z) = 0).
38
7 Complex functions
The aim here is to take a complex function w = f(z) (meaning a complex-valued function taking complex arguments, examples are w = z2 + z − 1, w =(z − 1)/(2z + i), w = cos(z2 + (3− 2i)z), w = e−5iz−1
), and try to understandits effect.
In the case of real-valued functions on the real line we can do this by sketchingthe graph of the function but to do this in the case of complex-valued functionsrequires four real dimensions (since the complex plane is already 2-dimensionalover the reals), which we don’t have at our disposal (though taking 3-dimensionalsections through time can be a reasonable substitute for four real dimensions).
What we do here is to take lines and curves in the z-plane (the domain ofthe function f) and see what (1-dimensional figures) they become in the (image)w-plane. By choosing various (systems of) lines and curves in the z-plane wecan build up a picture of the effect of the function.
Example 7.1 w = f(z) = z2
First consider a (vertical) line L parallel to the imaginary axis, say theequation of L is x = x0 (where x0 is a fixed real constant). Thus L consists ofall complex numbers of the form z = x0 + iy with y ∈ R.
Now,w = f(x0 + iy) = (x0 + iy)2 = (x2
0 − y2) + 2ix0y.
Here x0 is fixed and y can be any real number.Set w = u + iv so u = x2
0 − y2 and v = 2x0y (*6). What figure do thesepoints u + iv form in the w-plane?
We have v2 = 4x20y
2 = 4x20(x
20−u) = 4x4
0− 4x20u, which is the equation of a
parabola opening in the negative real direction and with vertex at (x20, 0) (and,
you can check, focus at 0).
39
//
OO
◦
What happens if we allow x0 to vary? In other words consider a family ofparallel lines, some of which are as shown. Here we’ve changed the notation sothat xa means the line x = a.
//
OO
x−2→
x−1→
x0→ x 1
2→
x1→
x 32→
x2→
The images of these are as shown.
40
//
OO
◦
x 14
= x− 14
x1 = x−1
x 32
= x− 32
x2 = x−2
The case x0 = 0 is “degenerate” in the sense that the parabola closes up intoa straight half-line. Notice that if x0 = 0 and y varies from negative to positive,the corresponding point in the image half-line runs in towards 0 and then outagain. Similarly for any non-zero value of x0, say x0 = 1: as y varies fromnegative to positive, the corresponding point on the parabola runs in from thebottom left and out towards the top left.
Thus, although the lines given by x0 = 1 and x0 = −1 map to the samecurve under the action of the function f , the points run in opposite directionsas y varies (you should check this for yourself using equation (*6)).
41
//
OO
TBCIC
We can also consider the effect of the function f(z) = z2 on the horizontallines: these have the form y = y0 where y0 is constant so, fixing y0, we fixsuch a horizontal line, namely that consisting of complex numbers of the formz = x + iy0.
Then w = f(z) = f(x + iy0) = (x + iy0)2 = (x2 − y20) + 2ixy0 (y0 fixed,
x ∈ R) = u + iv where u = x2 − y20 and v = 2xy0.
In order to understand what curve in the u, v-plane these equations repre-sent we proceed as in the case of vertical lines, but this time eliminating the“parametrising variable” x:
v2 = 4x2y20 = 4(u + y2
0)y20 = 4uy2
0 + 4y40
which is the equation of a parabola opening in the positive real direction.
To see why f(z) = z2 pulls these vertical and horizontal lines into curves,fix x0 and look at the effect of f on the vertical line z = x0 + iy:
f(x0 + iy) = (x20 − y2) + 2ix0y
//
OO
/x0
◦x0 + iy
The modulus of f(x0 + iy) is
42
√(x2
0 − y2)2 + (2x0y)2 =√
x40 − 2x2
0y2 + y4 + 4x2
0y2 =
√x4
0 + 2x20y
2 + y4 =√(x2
0 + y2)2 = x20 + y2 = |x0 + iy|2
so one effect of f is to square the distance of points from the origin.
(Of course it is easier to see this if we think in terms of polar coordinates:
f(reiθ) = (reiθ)2 = r2ei2θ.
So the total effect of f on a complex number z is to square its modulus androtate it anticlockwise by an angle equal to the argument of z.)
//
OO
/ x0
kkkkkkkkkkkkkkkkkkk
◦r
zzzzzzzzzzzzzzzzzzzzzzz /x1
◦
θ TBCIC
With the notation of the diagram we have cos(θ) = x0/r and cos(2θ) =x1/r2. Now cos(2θ) = 2cos2(θ) − 1 = 2x2
0/r2 − 1 so x1 = 2x20 − r2 which is
strictly less than x20. On the diagram I have taken x0 = 1 so that x2
0 = x0 inorder to illustrate the “bending” of the line directly.
Thinking of the effect in this way perhaps makes it clearer how the functionf pulls the straight line z = x0 + iy into a curve.
The above analysis suggests that thinking in terms of polar coordinatesz = reiθ would also help us to understand the effect of this squaring function.So let’s consider what happens to the lines θ = a constant
//
OO
ooooooooooooo
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and circles r = a constant
43
//
OO
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First, fix θ0 and consider the radial half-line z = reiθ0 , r ∈ R, r > 0 (whichdoes not include the origin, but it’s obvious that 0 is fixed by the squaringfunction). Then f(z) = f(reiθ0) = r2ei2θ0 - so this line is rotated anticlockwisethrough an angle of θ0 and the distances from the origin are squared - thereforethe point on this line at distance 1 from 0 stays at distance 1 (so is just rotated)points at distance greater than 1 are moved further out, and points at distanceless than 1 are moved closer to 0.
Note that the line with θ0 = 0 is fixed whereas if θ0 is just below 2π then thecorresponding line is rotated almost twice, yet these two lines are close together:there appears to be a discontinuity here and we will come back to this a littlelater.
//
OO
//
OO
//
OO
TBCIC
Now, instead of fixing the angle θ, fix r0 and consider the circle z = r0eiθ
θ ∈ [0, 2π). We have f(z) = (r0eiθ)2 = r2
0ei2θ and this is the equation of a circle,
still centred at 0, but with radius r20.
44
Also note that the mapping
r0eiθ 7→ r2
0ei2θ
wraps the original circle twice round the image circle, because the argument isdoubled: 2θ ∈ [0, 4π).
Putting all this partial information together gives a fair idea of the effectof the squaring function. However, using a surface intermediate between thez-plane and the w-plane can clarify the situation more.
We’ve seen that the effect of the squaring function is to square moduli andalso to wrap the plane twice round itself, so we can introduce a surface which,unlike the plane itself, can show that we have to wrap twice before getting backto where we started.
The surface we want to use cannot be embedded in 3-dimensional space: itneeds 4 dimensions (really it is the graph of f , so needs 2× 2 dimensions) butif we allow it to cut through itself then we can almost get an embedding of itin 3-space.
I will give a picture of this surface on a separate sheet (it would take me along time to produce the pictures using the package I am using for diagramswithin these notes). Here are a couple of links to pictures:
http://www- math.mit.edu/18.04/18.04-rrr/Pictures andhttp://mathworld.wolfram.com/RiemannSurface.html
D1 - this is a reference to a diagram on a separate sheet (similar referencesto separate diagrams appear below)
So the action of f can be broken into two stages: map the z-plane to theRiemann surface (which is really the graph of f) and then project to the w-plane.
The analogous process in the case of a real function of one variable wouldbe to map points on the x-axis to the corresponding points on the graph of thefunction and then map across to the y-axis.
//
OO
like z-plane
like w-plane
This Riemann surface gives a nice picture of the square root function - the“inverse” of the squaring function (not really an inverse since it is many-valued,for instance
√4 = ±2 and
√−1 = ±i). The process of taking a square root of a
45
complex number can be thought of as: given w ∈ C choose w′ on the Riemannsurface such that πw′ = w, where π is the projection of the Riemann surfaceto the w-complex plane. Clearly for w 6= 0 there are two choices (one, w′′, onthe “upper” (only as drawn) and one on the “lower” part of the ramp). Theexception is w = 0, which has just one square root.
D2 - this is a reference to a diagram on a separate sheet
This also illustrates the fact that there is no way to choose a square root ofa complex number in a “continuous” way (see the comment above about linesclose together being rotated very different amounts).
For, take a circle which has the point 0 in its interior and try to lift thisin a continuous way to the Riemann surface - that is, try to make a continouschoice of square root - you can see that this is not possible: the start and endpoints on the “lifted circle” are distinct (one directly above the other). On theother hand, note that one can lift, in a continous way, any circle which does notcontain the origin.
D3 - this is a reference to a diagram on a separate sheet
Example 7.2 f(z) = z3
We won’t go through this in any detail but note that, writing z = reiθ, wehave f(z) = r3ei3θ so the effect of this cubing function on, say, the unit circle isto wrap it round itself three times so the corresponding Riemann surface, whichis essentially the graph of f, is a triple ramp.
Example 7.3 f(z) = zn
Here one has an n-fold ramp for the Riemann surface.
Example 7.4 f(z) = (z − 1)2
This is very like f(z) = z2 but with the origin z = 0 replaced by z = 1.
8 The exponential and logarithm functions
The “graph” of the exponential function f(z) = ez is a rather more complicatedsurface than those we saw in the previous section.
First consider the effect of this function on a horizontal line z = x + iy0. Wehave f(z) = exeiy0 : this is the ray in the w-plane containing eiy0 (and omitting0) with 0 + iy0 going to the point eiy0 on the unit circle in the w-plane.
46
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z-plane
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//
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??
w-plane
__________ ◦
??
??
?
◦
Now for the fun: consider the effect on a vertical line z = x0 + iy. Thenf(z) = ex0eiy - this is a complex number in the w-plane of modulus ex0 butas y varies the image of the line is wrapped around the circle C(0, ex0) in thew-plane infinitely many times.
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//
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??
z-plane
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//
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??
w-plane
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So the intervening surface needs infinitely many turns of the ramp (in bothdirections) and never closes up in the way that the surfaces corresponding to thefunctions f(z) = zn do. You can find a picture at http://www-math.mit.edu/18.04/18.04-rrr/Pictures .
47
Look at the effect on the line z = iy for instance.������������������������� �������������������������
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..........
Recall that for real numbers u, v the logarithm function is defined by
v = eu iff logev = u.
This makes sense because for each value of v there is exactly one value of uwith eu = v. However, in the case of complex numbers we have just seen thatfor each non-zero complex number w there are infinitely many values of z withez = w. So there is no unambiguously defined logarithm function on complexnumbers.
It is, however, useful to be able to reverse the action of the exponentialfunction at least locally, so we just accept that complex logarithm is multivalued(and, if we like, we can pick a particular range to make it single-valued - justas in the definition of the inverse trigonometric functions sin−1 (= arcsin) etc.)but one has to accept then that such a “principal value of log” is not continuouseverywhere (just as there is no continuous way to make a choice of square rootsof complex numbers, or make a continuous choice of sin−1, there is no way tomake a continuous choice of log).
So we setlog(w) = z iff ez = w.
To make this more explicit, set w = seiφ and z = x+ iy so seiφ = w = exeiy.Hence s = ex so x = loges (since both s and x are real this is unambiguous)and φ = y + 2kπ.That is, x + iy = log(w) iff x = loge|w| and y = arg(w) (remembering thatarg(w) is multi-valued: any chosen value +2kπ for k ∈ Z). That is,
log(w) = log|w|+ iarg(w).
48
If we write w in polar form, as w = reiθ, this becomes
log(w) = log(r) + iθ(+2kπ).
One can, if one wishes, choose a principal value of log, say by insistingthat arg(w) ∈ [0, 2π).
49
Example 8.1 log(i) = log|i|+ iarg(i) = 0 + i(π2 + 2kπ)
So a couple of the values of log(i) are π2 i, 5π
2 i.
//
OO
◦
◦
◦
This now allows us to define arbitrary exponentials of complex numbers; thatis, to make sense of zz′ where z and z′ are any complex numbers. Usually thiswill be many-valued. The results are quite interesting: for example, probably itis not immediately obvious what the value of ii is!
If w = zz′ then, taking (multivalued) log of each side, we havelog(w) = log(zz′) = z′log(z) (the usual rules for log are still valid in this con-text). Here “=” really means “the possible values on one side are the same asthe possible values on the other side”.
Taking exponentials gives
w = elog(w) = ez′log(z),
that is,zz′ = ez′log(z) = ez′(log|z|+iarg(z))
(usually multivalued)
Example 8.2 2i = eilog(2) = ei(log(2)+i2kπ) = e−2kπeilog(2) (first “log” meansreal log, second “log” means complex, hence many-valued, log). All these valueslie on a single ray from the origin. This ray contains the value obtained atk = 0, namely eilog(2) which, being of the form eiθ, is a complex number on theunit circle and, since log(2) ≈ .693 is in roughly the position shown.
//
OO
_ekr{����� % + 3:DLSY _ ekrz�����%+2:DLSY
•
50
Example 8.3 ii = eilog(i) = ei(i( π2 +2kπ)) = e−
π2−2kπ = e−
π2−2kπ. All these
values are real and a few are shown, not at all to scale, below.
//
OO
_ekr{����� % + 3:DLSY _ ekrz�����%+2:DLSY
• • •••
51
9 Continuity and differentiability
As with functions taking real arguments and values, we say that a functionf : C −→ C is continuous at z0 ∈ C if f(z) → f(z0) as z → z0. More precisely,choose ε > 0 and consider the ε-neighbourhood of f(z0):
Bε(f(z0)) = {w : |w − f(z0)| < ε}.
(Note that the boundary of this open disc is the circle, C(f(z0), ε), with centref(z0) and radius ε.)The question is then whether there is a (small) neighbourhood of z0 which ismapped into Bε(f(z0)) by f .
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◦f(z0)
w-planeThat is, given ε > 0 is there δ > 0 such that
fBδ(z0) ⊆ Bε(f(z0))
whereBδ(z0) = {z : |z − z0| < δ}?
Roughly then, f is continuous at z0 if we can ensure that f(z) is close tof(z0) by choosing z close enough to z0.
Most of the functions we will see are continuous at every, or almost every,point where they are defined. For example, polynomial functions, ez, trigono-metric functions and hyperbolic functions are continuous at every point. If fand g are continuous (at z0) then so are f + g and f × g, and so also is f/g ex-cept at those points where g = 0 (since then f/g is undefined (unless “f equals
52
0 to at least as high a degree as g”)). We will sometimes say that f/g has asingularity at such a point where it is undefined.
If f is continuous at z0 then we can follow the real case and tentatively definethe derivative of f at z0 to be the limit of the fraction
f(z)− f(z0)z − z0
as z approaches z0, if this limit exists. That is, we set
f ′(z0) = lim|z−z0|→0f(z)− f(z0)
z − z0.
We can be more precise about what we mean by lim|z−z0|→0 and use the defi-nition
f ′(z0) = limr=|z−z0|→0,θ∈[0,2π)f(z0 + reiθ)− f(z0)
(z0 + reiθ)− z0.
Note that the complex numbers z0 + reiθ are those right on the boundary ofthe disc, Br(z0), of radius r around z0. So the formula looks at the values of fon this circle, subtracts the value, f(z0), at the centre of the circle, and dividesby the complex number z − z0 (which is of modulus r).
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• ◦
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◦
z0z0 + rei0
z0 + rei π4
The question is then whether the resulting complex numbers tend to a defi-nite limit as θ varies and as r → 0 (that is, as the circle shrinks towards z0). Ifthey do, then this common value, f ′(z0), is the derivative of f at z0 and theresulting function is (where defined) the derivative, f ′, of f .
This is a limit that is much more difficult to achieve compared with the real-valued situation: imagine fixing θ1, setting z = z0 + reiθ1 , and then allowingr to approach 0 - so the point z approaches z0 “from the direction θ1”. Onemight well imagine that the limit
limr=|z−z0|→0f(z = z0 + reiθ1)− f(z0)
z − z0
could exist for each choice of θ1 but that the value of the limit would dependon the angle θ1 chosen (in which case the derivative would not be defined).
Compare with the somewhat analogous situation where we have a functiong : R2 −→ R. If g is differentiable at a point then the derivatives in variousdirections can be put together to form a tangent plane (to g at that point) butthis plane may be tilted at any angles - in particular, two vectors are needed todetermine this tangent plane.
53
The complex situation is more rigid: a single vector is enough to determinethe tangent plane (because the tangent plane is one-dimensional as a complexvector space, whereas in the case of g : R2 −→ R the tangent plane is a two-dimensional real vector space).
Roughly, “complex-differentiable = real-differentiable +C-linear”.
The Cauchy-Riemann equations below will provide an easily checkable crite-rion for deciding whether or not a complex function is differentiable at a pointand they also give an expression for the derivative. The rough idea is that weregard f : C −→ C temporarily as a function from R2 −→ R2, we check thatthe partial derivatives of this real function exist and then the Cauchy-Riemannequations give the extra requirement for f to be differentiable as a function ofa complex variable. In fact, we broaden the context by considering functionswhich are defined on a region of C, meaning an open (containing no boundarypoints) connected subset of C.
Complex-differentiability of a function has very strong consequences: thederivative of a complex-differentiable function is itself complex-differentiableand so, therefore, all the derivatives of the function exist. It is a consequenceof this that if f is complex-differentiable in a neighbourhood of a point a thenthere is a power series expansion of f around a, as
f(z) =∞∑
n=0
f (n)(a)n!
(z − a)n.
We will also see that a function being complex-differentiable near a point isessentially equivalent to the function having a power-series expansion at thatpoint.
To derive the Cauchy-Riemann equations, we can proceed as follows.Given the function f : C −→ C set w = f(z) and write w = u + iv and
z = x+ iy. So u+ iv = f(x+ iy) and hence u and v can be thought of as u(x, y)and v(x, y), that is, as real-valued functions of two real variables.
Example 9.1 If f is the squaring function f(z) = z2 then u+ iv = (x+ iy)2 =(x2 − y2) + 2ixy so u = x2 − y2 and v = 2xy.
If f is complex-differentiable then, it can be checked, this implies that allfour partial derivatives
∂u
∂x,
∂u
∂y,
∂v
∂x,
∂v
∂y
exist. Also the usual rules for calculating derivatives, in particular the chainrule, hold (because they are more or less formal consequences of the definitionof derivative) so, since w = u + iv we have
∂w
∂x=
∂u
∂x+ i
∂v
∂xand
∂w
∂y=
∂u
∂y+ i
∂v
∂y
54
but also, by the chain rule, since w is a function of z and z is a function of x wehave
∂w
∂x=
dw
dz
∂z
∂x
(we write dwdz not ∂w
∂z because there we’re regarding w as a function of just onevariable z)
=dw
dz· 1 =
dw
dz.
Similarly∂w
∂y=
dw
dz
∂z
∂y=
dw
dzi
(since z = x + iy)Putting these pairs of equations together gives
dw
dz=
∂w
∂x=
∂u
∂x+ i
∂v
∂x
andidw
dz=
∂w
∂y=
∂u
∂y+ i
∂v
∂yso
dw
dz= −i
∂u
∂y+
∂v
∂y.
This gives∂u
∂x+ i
∂v
∂x=
∂v
∂y− i
∂u
∂y
so, equating real and imaginary parts,∂u∂x = ∂v
∂y
∂v∂x = −∂u
∂y
These are the Cauchy-Riemann equations.
Example 9.2 Continuing with the squaring function, we have
∂u
∂x= 2x and
∂v
∂y= 2x (equal),
also∂u
∂y= −2y and
∂v
∂x= 2y (negatives of each other)
so the squaring function f(z) = z2 does satisfy the Cauchy-Riemann equations.
55
We have argued (not very rigorously but a more careful argument is givenlater) that if the function f is complex-differentiable then f must satisfy theCauchy-Riemann equations. In fact, if all four partial derivatives appearingare defined and continuous then the converse is true, so we have a checkablecriterion for complex-differentiability.
Theorem 9.3 Given a complex function f defined on a region, define the real-valued functions, u and v, of two variables by f(x+iy) = u(x, y)+iv(x, y). Sup-pose that all four partial derivatives u and v exist and are continuous. Then f iscomplex-differentiable iff these partial derivatives satisfy the Cauchy-Riemannequations.
In fact this is true locally: if z0 ∈ C and if the four partial derivatives aredefined and continuous in a neighbourhood of z0 then f is complex-differentiableat z0 iff the Cauchy-Riemann equations hold at z0.
In this case, the derivative of f (at z0) is
f ′ =∂u
∂x+ i
∂v
∂x
(=
∂v
∂y− i
∂u
∂y
).
Example 9.4 f(z) = cos(z)
Write z = x + iy so u + iv = cos(x + iy) = cos(x)cos(iy)− sin(x)sin(iy)= cos(x)cosh(y)− isin(x)sinh(y).
Therefore u = u(x, y) = cos(x)cosh(y) andv = v(x, y) = −sin(x)sinh(y)
Compute the four partial derivatives:
∂u
∂x= −sin(x)cosh(y) and
∂v
∂x= −cos(x)sinh(y)
∂u
∂y= cos(x)sinh(y) and
∂v
∂y= −sin(x)cosh(y)
and note that we do have
∂u
∂x=
∂v
∂yand
∂u
∂y= −∂v
∂x.
that is, the Cauchy-Riemann equations are satisfied.
So, since the conditions on the partial derivatives (being defined and contin-uous) are satisfied, we conclude that f(z) = cos(z) is a complex-differentiablefunction and has derivative
−sin(x)cosh(y) + i(−cos(x)sinh(y)) = −(sin(x)cosh(y) + cos(x)sinh(y)
)which, you can check, equals −sin(z).
In fact, all the “usual” polynomial, trigonometric, hyperbolic, exponentialfunctions are complex-differentiable.
56
Example 9.5 The function f(z) = z, which may be regarded as reflecting thecomplex plane in the real axis.
Write z = x + iy so u + iv = f(x + iy) = x− iy.Hence u = u(x, y) = x and v = v(x, y) = −y (so u does not depend on y
and v does not depend on x). Then
∂u
∂x= 1 and
∂v
∂x= 0
∂u
∂y= 0 and
∂v
∂y= −1.
So we do not have ∂u∂x = ∂v
∂y at any point and we conclude that complexconjugation function is nowhere differentiable.
Here is a direct argument, working just from the definition, showing thatthe Cauchy-Riemann equations necessarily hold if f is complex-differentiable.
Assume that f is differentiable, so at every point, z0 = x0 + iy0, the limit
f ′(z0) = limz→z0
f(z)− f(z0)z − z0
exists.Set h = z − z0 and note that
f ′(z0) = limh→0f(z0 + h)− f(z0)
h= limh→0
f(x0 + iy0 + h)− f(x0 + iy0)h
.
Here h is a complex number, approaching 0, and we will first constrain h toapproach 0 along the real axis. Set z = x + iy so u and v are functions of x andy. First we have, with h a real variable,
f ′(z0) = limh→0,hu(x0 + h, y0) + iv(x0 + h, y0)− u(x0, y0)− iv(x0, y0)
h
= limh→0,hu(x0 + h, y0)− u(x0, y0)
h+ ilimh→0,h
v(x0 + h, y0)− v(x0, y0)h
=∂u
∂x
∣∣∣∣(x0,y0)
+ i∂v
∂x
∣∣∣∣(x0,y0)
(by definition of partial derivatives).Similarly, constrain h instead to approach 0 along the imaginary axis, so
h = ik for k real and z0 + h = x0 + iy0 + ik = x0 + i(y0 + k). Then
f ′(z0) = limk→0,ku(x0, y0 + k) + iv(x0, y0 + k)− u(x0, y0)− iv(x0, y0)
ik
57
= limk→0,ku(x0, y0 + k)− u(x0, y0)
ik+ ilimk→0,k
v(x0, y0 + k)− v(x0, y0)ik
=1ilimk→0,k
u(x0, y0 + k)− u(x0, y0)k
+i
ilimk→0,k
v(x0, y0 + k)− v(x0, y0)k
=1i
∂u
∂y
∣∣∣∣(x0,y0)
+∂v
∂y
∣∣∣∣(x0,y0)
=∂v
∂y
∣∣∣∣(x0,y0)
− i∂u
∂y
∣∣∣∣(x0,y0)
.
In each case the limit is f ′(z0) (which we’re assuming exists) so, equatingreal and imaginary parts, we deduce that
∂u
∂x=
∂v
∂yand
∂v
∂x= −∂u
∂y
at z0 = x0 + iy0.
The argument also shows that
f ′(z0) =∂u
∂x+ i
∂v
∂x
(=
∂v
∂y− i
∂u
∂y
).
And here, in detail, is the point about real tangent planes versus complextangent lines.
With the usual notation, u + iv = w = f(z = x + iy), think of f as afunction (x, y) 7→ (u(x, y), v(x, y)) from R2 to R2. Now, if this is differentiableas a function of real variables then all directional tangent vectors will lie on asingle plane, which is spanned by, for example, the vectors:(∂u
∂x,∂v
∂x
)(tangent vector in x-direction) and
(∂u
∂y,∂v
∂y
)(tangent vector in y-direction).
As complex numbers (that is, thinking of the Argand diagram rather than thereal plane) these are
∂u
∂x+ i
∂v
∂xand
∂u
∂y+ i
∂v
∂y
If the function is complex-differentiable then i is a scalar so it will make nodifference whether we multiply x + iy by i first and then differentiate, or dif-ferentiate and then multiply by i (since f ′(cz) = cf ′(z) if c is a constant).Multiplying first by i has the effect of rotating the complex plane anticlockwisethrough π
2 and, therefore, this will rotate the positive real tangent vector, inwhose direction the derivative is
∂u
∂x+ i
∂v
∂x
58
to the positive imaginary direction, in whose direction the derivative is
∂u
∂y+ i
∂v
∂y.
Therefore the latter vector should be i times the former vector, that is:
∂u
∂y+ i
∂v
∂y= i
(∂u
∂x+ i
dv
∂x
)= i
∂u
∂x− ∂v
∂x
so, equating real and imaginary parts, we obtain
∂u
dx=
∂v
∂yand
∂v
∂x= −∂u
∂y,
that is, the Cauchy-Riemann equations, as required.Note the following re-phrasing of Theorem 9.3.
Theorem 9.6 If u and v are real-valued functions of two real variables x, ywith continuous partial derivatives which satisfy the Cauchy-Riemann equationsare satisfied then the complex function f defined by f(x+ iy) = u(x, y)+ iv(x, y)is complex-differentiable.
Note also that the functions u and v above satisfy Laplace’s equation∇2 = 0:
∇2u =∂2u
∂x2+
∂2u
∂y2
=∂
∂x
(∂u
∂x
)+
∂
∂y
(∂u
∂y
)=
∂
∂x
(∂v
∂y
)+
∂
∂y
(− ∂v
∂x
)=
∂2v
∂x∂y− ∂2v
∂y∂x=
∂2v
∂x∂y− ∂2v
∂x∂y= 0
and similarly for v.
Example 9.7 Find all points where the function f(z) = |z| is differentiable andfind the derivatives at any such points.
Set z = x + iy, so u + iv = f(x + iy) = (x2 + y2)12 . Hence
∂u
∂x=
12 · 2x
(x2 + y2)12
(provided z 6= 0) whereas ∂v∂y = 0 and so these can be equal only if x = 0.
Similarly ∂u∂y = − ∂v
∂x only if y = 0.
59
Therefore the only point where f could possibly be differentiable is z = 0.There, however, the expression,
x
(x2 + y2)12,
for ∂u∂x is not valid and even computing
limx→0x
(x2 + y2)12
in order to compute ∂u∂x at 0 will be valid only if ∂u
∂x is a continuous function at0. At this point we should remember that even the real function f(x) = |x| isnot differentiable at 0 (from the left it has slope −1 but, from the right, slope+1) so certainly its extension to the complex plane cannot be continuous at 0.
We conclude that f(z) = |z| is nowhere differentiable.
Suppose that f is complex-differentiable. If z0 is any point where the deriva-tive f ′(z0) 6= 0 then, it can be shown, the “local” effect of f is a combinationof translation, scaling and rotation. In particular, f is “angle- preserving” atsuch points. For example, we saw that the squaring function f(z) = z2 takesthe horizontal and vertical coordinate lines into curves but you can check thatthese curves meet at right angles, just as the horizontal and vertical lines did.The exception is the point z0 = 0, where the function is not angle-preserving,indeed it doubles angles around 0: but that just illustrates why the statementabove is for points where the derivative is non- zero.
The Cauchy-Riemann equations indicate why this is so: the derivative off = u + iv in the x-direction is
∂u
∂x+ i
∂v
∂x
and in the y-direction is∂u
∂y+ i
∂v
∂y
and these two directions are perpendicular: for consider the dot product(∂u
∂x,∂v
∂x
)·(
∂u
∂y,∂v
∂y
)=
∂u
∂x· ∂u
∂y+
∂v
∂x· ∂v
∂y=
∂u
∂x· −∂v
∂x+
∂v
∂x· ∂u
∂x= 0.
So the perpendicular x- and y- tangent directions are transformed by f toperpendicular directions. Therefore, locally, f “rotates (tangent) directions”and hence preserves angles between (tangent) directions.
60
10 Taylor and Laurent expansions: singularitiesand poles
If f is a complex function all of whose derivatives exist and if a is a complexnumber we can write down the Taylor expansion of f around a, just as for realfunctions.
f(z) = f(a) + (z − a)f ′(a) +(z − a)2
2!f ′′(a) + · · ·+ (z − a)n
n!f (n)(a) + . . .
=∞∑
n=0
(z − a)n
n!f (n)(a)
where f (n) denotes the n-th derivative of f .
Just as for real power series there is the issue of convergence: in general,given a complex-differentiable function f and a complex number a there will bea largest value R (the radius of convergence), which will be a non-negativereal number or “∞”, such that for complex numbers z with |z − a| < R thepartial sums
N∑n=0
(z − a)n
n!f (n)(a)
tend to a limit, which we write as∞∑
n=0
(z − a)n
n!f (n)(a),
and which will be equal to f(z). (And it is the case that for z with |z − a| > Rthe series diverges and, as you might expect from the real case, the situation onthe boundary, |z| = R, can be complicated.) In fact we have the following.
Theorem 10.1 If a ∈ C and r ∈ R with r > 0 and if f is complex-differentiableat every point of the closed disc Br(a) = {z : |z − a| ≤ r} then at every point zin the open disc Br(a) = {z : |z − a| < r} the Taylor expansion
f(z) =∞∑
n=0
(z − a)n
n!f (n)(a)
is valid.(It is also the case that for such z one can differentiate term-by- term:
f ′(z) =∞∑
n=1
(z − a)n−1
(n− 1)!f (n)(a).
Indeed, for such z one can also integrate term-by-term: we discuss integrationlater.)
61
One can check that all the previous power series expansions, valid for realnumbers, are valid for complex numbers within the limits, if any, indicated.
ez = 1 + z +z2
2!+
z3
3!+ . . .
sin(z) = z − z3
3!+
z5
5!− . . .
cos(z) = 1− z2
2!+
z4
4!− . . .
sinh(z) = z +z3
3!+
z5
5!+ . . .
cosh(z) = 1 +z2
2!+
z4
4!+ . . .
(1 + z)n = 1 + nz +n(n− 1)
2!z2 +
n(n− 1)(n− 2)3!
z3 + . . . |z| < 1
log(1 + z) = z − z2
2+
z3
3− . . . |z| < 1
In the expansion of (1 + z)n we take n to be any integer. Particular cases are:
11 + z
= 1− z + z2 − . . . |z| < 1
11− z
= 1 + z + z2 + . . . |z| < 1.
Of course, if n is a non-negative integer, the expansion of (1+z)n is actually validfor all z because it is really just a finite expansion (the product n(n−1)(n−2) . . .is eventually 0). If n is any other real or complex number then some caution mustbe exercised since the function is many-valued so is well-defined, continuous anddifferentiable only locally around z - in particular 0 cannot be included in anyregion where it is single-valued. A similar comment applies to log(1 + z).
We want to have similar expansions for functions with singularities. Forexample the function z/(1− z) is defined, continuous and differentiable exceptat the point z = 1, where it is undefined, or “goes off to infinity”. We will referto points where a function f(z) is not well-defined as singularities of f .
For example,
f(z) =1
z2 + 1=
1(z − i)(z + i)
has two singularities, one at i, the other at −i.
Suppose that f has a singularity at z = a. If m is the least positive integersuch that (z − a)mf(z) is well-defined at a then we will say that f has a poleof order m at a. A pole of order 1 is also referred to as a simple pole (andone of order 2 is a double pole etc.). (If there is no such m then we have asingularity which is not a pole.)
62
Example 10.2
f(z) =1
z2 + 1The function f is not even defined at z = i but
(z − i)f(z) =1
z + i
is defined and differentiable at z = i so
1z2 + 1
has a pole of order 1 (a simple pole) at i.
Example 10.3 If
f(z) =z2
(z − i)4
then, at z = i, none of f(z), (z− i)f(z), (z− i)2f(z), (z− i)3 = f(z)z2/(z− i) iswell-defined but (z− i)4f(z) = z2 is well-defined and differentiable so z2/(z− i)4
has a pole of order 4 at z = i.
Example 10.4 Take f(z) = e1/z: this is undefined at z = 0. For any positiveinteger m, (z − 0)mf(z), that is zme1/z, is still undefined: for
zme1z = zm
(1 +
1z
+1
2!z2+
13!z3
+ . . .)
and the zm cannot cancel out the singularity produced by the terms 1/zn for nlarger than m.
Therefore the singularity of e1/z is not a pole; it is an example of an essentialsingularity, by which we will mean a non-removable singularity of a functionwhich is not a pole.
In order to deal with functions which have singularities we will consider serieswith possibly some negative powers of z − a. In the example above we saw onesuch series, the expansion of e1/z about z = 0, with infinitely many negativepowers of z(−0). Expressions with only finitely many negative powers of z − aare, in contrast, quite well-behaved.
An expression of the form∑∞
n=K an(z − a)n where K is a possibly negativeinteger, will be called a Laurent series (so, if K ≥ 0 this is a Taylor series).
The part of a Laurent series involving negative exponents,
−1∑n=K
an(z − a)n,
63
is called the principal (or singular) part. If the Laurent series defines afunction f by the formula f(z) =
∑∞n=K an(z − a)n for all z 6= a in some disc
round a then we will refer to this as a Laurent expansion of f at a and thecoefficient, a−1, of (z − a)−1 is called the residue of f at a, written res(f, a).
Example 10.5 Let
f(z) =ez
z4.
Expanding, we have
ez
z4=
1z4
(1 + z +z2
2!+ · · ·+ zn
n!+ . . . )
= z−4 + z−3 +12z−2 +
13!
z−1 +14!
(z0) +15!
z1 + . . .
and this is the Laurent series expansion of (ez)(z4) round z = 0.The principal part of this expansion is
z−4 + z−3 +12z−2 +
13!
z−1
and the residue of ez/z4 at 0 is 1/3! = 1/6.
Example 10.6 Let
f(z) =ez
(z − 1)4.
To expand this as a Laurent series around z = 1 we need to expand ez inpowers of z − 1:
ez
(z − 1)4=
e1+(z−1)
(z − 1)4=
e1ez−1
(z − 1)4=
=e
(z − 1)4(1 + (z − 1) +
(z − 1)2
2!+ . . .
)and then continue as in the previous example.
Example 10.7 To expand1
z2 + 1around i, write
1z2 + 1
=1
(z − i)(z + i)=
1z − i
· 1(z − i) + 2i
=
=1
z − i· 12i(1 + z−i
2i
) =
64
=12i· 1z − i
(1− z − i
2i+(z − i
2i
)2
−(z − i
2i
)3
+ . . .
)(using that 1
1+w = 1− w + w2 − w3 + . . . )
=12i
(z − i)−1 − 1−4
+1
8(−i)(z − i)− 1
16(z − i)2 + . . .
=−i
2(z − i)−1 +
14
+i
8(z − i)− 1
16(z − i)2 + . . .
In particular,
res(
1z2 + 1
, i
)= − i
2
If f has a simple pole at z = a then we can use either of the following quickmethods to compute res(f, a).
(1) If
f(z) =g(z)z − a
where g(a) 6= 0 then res(f, a) = g(a).(Proof: g is complex-differentiable so has a power series expansion round a:g(z) = b0 + b1(z − a) + b2(z − a)2 + . . . and g(a) 6= 0 implies b0 6= 0. Then theLaurent series expansion of f round a is
f(z) =b0
z − a+ b1 + b2(z − a) + . . .
so res(f, a) = b0 = g(a).)
For example if
f =z3
z2 + 1=
z3
(z − i)(z + i)
then, to calculate the residue of f at z = i:
f =z3/(z + i)
z − i
so res(f, i) = i3/(2i) = −1/2.
(2) If
f(z) =g(z)h(z)
has a simple pole at z = a and if g(a) 6= 0 then
res(f, a) =g(a)h′(a)
.
65
(Proof: the Taylor expansion of h at z = a is
h(z) = h(a)+h′(a)(z−a)+h′′(a)
2!(z−a)2+· · · = h′(a)(z−a)+
h′′(a)2!
(z−a)2+. . .
sog(z)h(z)
=g(z)
(z − a)(h′(a) + (z − a)
(h′′(a)2! + . . .
)) .
Therefore, by (1) above with
g(z)
h′(a) + (z − a)(h′′(a)
2! + . . .)
replacing g(z) there, we have
res(f, a) =g(a)
h′(a) + 0=
g(a)h′(a)
.)
For example, take f as above: then
res(f, i) =i3
2z|z=i=
i3
2i=−12
.
11 Integration
Suppose that f is a complex function defined in a region of the complex planeexcept possibly for finitely many points in that region, where it has poles. Sup-pose also that C is a curve, a “contour”, of some sort in the region where f isdefined and which is such that none of the poles of f lies on C. In this sectionwe will define the integral of f along C, denoted
∫C
f or∫
Cf(z)dz. Then we will
give a variety of methods of evaluating such integrals and various applicationsof such (“contour”) integration.
The idea seen in the context of real (Riemann) integration applies here aswell:
∫C
f(z)dz should be the “infinite sum of values of f(z) at points z on C,times “infinitely small” segments, dz, of the path C”. Also, just as in the realcase, we integrate along directed curves (recall that
∫ b
af = −
∫ a
bf : the first is
integrating along the directed path from a to b, the second along the directedpath from b to a).
So, if C looks like
66
approximate C by straight line segments, say ∆1,∆2, . . . ,∆8
LLLLee
∆1
�������
LL
∆2
xxxxxxxxxx
;;
∆3 BBBBBBBB
!!∆4
$$$$$$$$$
��
∆5
VVVVV**
∆6
��������
FF
∆7
DD
∆8
Each straight line segment ∆i should be thought of as a vector (roughly,a tangent vector or a chord to the curve) so can be thought of as a complexnumber in the complex plane.
LLLLee∆1
�������
LL ∆2xxxxxxxxxx
;; ∆3
BBBBBBBB
!! ∆4
etc.
Pick one point, zi, on each segment ∆i and evaluate f at zi.Then
∑81 f(zi)∆i (∆i thought of as a complex number) is a (very rough)
approximation to∫
Cf(z)dz.
By making the line segments smaller and smaller we should obtain betterand better approximations to
∫C
f(z)dz. More precisely, we define the integral∫C
f(z)dz to be the limit of these finite sums as the lengths of the line segmentsgo to zero (if this limit exists and is independent of the particular choices, ofline segments and “evaluation points”, made). Well, to define this properly onemust be a little more careful but this is pretty much the idea.
Example 11.1 Let C(0, 1) be the circle in the complex plane with centre 0 andradius 1. Let us evaluate
∫C(0,1)
z2dz. Actually this is ambiguous: we need tochoose an orientation of the circle, so let’s make anticlockwise orientation ourdefault (clockwise orientation would then give the negative of the number weobtain with the anticlockwise orientation).
We can approximate C(0, 1) by a regular polygon with n sides, as shown.
67
//
OO
$$$$RR
////WW
????__
OOOOgg
iiiiii
2πn
SS
Let’s take n = 4: a very coarse approximation but it will illustrate the idea.
//
OO
?????????????????
__
�����������������
��
?????????????????
��
�����������������
??
∆1∆2
∆3 ∆4
As vectors, ∆1, . . . ,∆4 are:∆1 = −1 + i and ∆2 = −1− i∆3 = 1− i and ∆4 = 1 + iLet’s pick the mid-point of each segment as the “evaluation point”:
z1 =12
+12i, z2 = −1
2+
12i, z3 = −1
2− 1
2i, z4 =
12− 1
2i
So
f(z1) =14(1+i)2, f(z2) =
14(−1+i)2, f(z3) =
14(−1−i)2, f(z4) =
14(1−i)2.
Therefore the corresponding approximation to the integral is
4∑1
f(zi)∆i =
=14
((1 + i)2(−1− i) + (1− i)2(−1− i) + (1 + i)2(1− i) + (1− i)2(1 + i)
)
68
= 0 (note that the terms cancel out in pairs).
In fact, if we choose a polygon with an even number of sides and choosediametrically opposite “evaluation points” then you can check that the termswill cancel out as in this example.
This suggests that∫
C(0,1)z2dz = 0 (it’s not a proof because we’ve not shown
that the result is independent of the segments and evaluation points chosen, butthis can be shown to be the case and 0 is, in fact, the right answer).
Example 11.2 Let C be the line from 0 to 1 + i and take f(z) = z2.
Divide the line into n equal pieces ∆1, . . . ,∆n so each is, as a complex num-ber, 1
n (1 + i)
//
OO
���������
??◦1 + i
0◦
and let’s choose the start-point of each segment for the evaluation point. Thatis,
z1 = 0, z2 = 1n (1+i), z3 = 2
n (1+i), . . . , zk = k−1n (1+i), . . . , zn = n−1
n (1+i)Then
n∑k=1
f(zk)∆k =n∑
k=1
(k − 1)2
n2(1 + i)2 · 1
n(1 + i)
=(1 + i)3
n3
n∑k=1
(k − 1)2 =(1 + i)3
n3
n−1∑k=0
k2
=(1 + i)3
n3· 16(n− 1)(n− 1 + 1)(2(n− 1) + 1)
=(1 + i)3
n3· 16(n− 1)(n)(2n− 1)
=(1 + i)3
6(1− 1
n)(1)(2− 1
n)
→ (1 + i)3
6· 2 =
13(1 + i)3 as n →∞.
Again, though we have not justified this rigourously, the integral∫
Cz2dz is
well-defined, so this is the correct value.
69
In the next section we will state some theorems which will allow us to calcu-late such integrals without having to resort to this kind of argument. But firstwe need to discuss parametrisation of curves in the complex plane.
By a contour we will mean a curve (containing its endpoints if it has any (i.e.if it isn’t a loop)) in the complex plane. Let γ be any contour. A parametri-sation of γ is a continuous function from an interval on the real line withimage γ: that is, φ : [a, b] −→ C such that γ = {φ(t) : t ∈ [a, b]} where[a, b] = {t ∈ R : a ≤ t ≤ b} if a ≤ b (if the curve crosses itself then some pointswill be parametrised by more than one value of t but that’s OK). Note that thisgives a direction/orientation to the curve if we think of t as moving from a tob. We will write [b, a] to mean the same interval but with t moving from b downto a so, if φ : [0, 1] −→ C is a parametrisation of C then φ : [1, 0] −→ C willmean this parametrisation of C but running in the opposite direction.
Example 11.3 To parametrise the line joining z1 to z2:z = (1− t)z1 + tz2 with t ∈ [0, 1]
(that is, φ : [0, 1] −→ C is defined by φ(t) = (1 − t)z1 + tz2; so note φ(0) = z1
and φ(1) = z2).
For instance, the line from 0 to 1 + i is parametrised as (1− t)0 + t(1 + i),which equals t(1 + i), for t ∈ [0, 1].
For instance, the line from −1 to 1 + i is (1− t)(−1) + t(1 + i) for t ∈ [0, 1],which can be rewritten as −1 + t + t + ti,= (2t− 1) + ti
//
OO
ooooooooooooo
77◦1 + i
−1◦
For instance the line from −i to i is parametrised as (1 − t)(−i) + ti =(−1 + 2t)i for t ∈ [0, 1].
Of course parametrisations are not unique, for example the last line abovemight more naturally be parametrised as ti for t ∈ [−1, 1].
Example 11.4 To parametrise the unit circle C(0, 1) with centre 0 and radius1, we may use z = eit with t ∈ [0, 2π] (that is, φ : [0, 2π] −→ C defined byφ(t) = eit).
Note that this is a closed curve (since any parametrisation by a closed intervalwill count the end-point twice).
70
Example 11.5 More generally the circle C(z0, r) with centre z0 and radius ris parametrised as z = z0 + reit for t ∈ [0, 2π].
//
OO
_ekr{����� % + 3:DLSY _ ekrz�����%+2:DLSY
◦z0 ◦t = 0
◦t = π4
◦t = π
2
To parametrise this in the other direction, clockwise, write z = z0 + re−it
for t ∈ [0, 2π], alternatively z = z0 + reit with t ∈ [2π, 0].
To parametrise the contour which turns round the first circle three timesanticlockwise, use z = z0 + reit for t ∈ [0, 6π], alternatively z = z0 + rei3t fort ∈ [0, 2π].
To parametrise the arc of the circle as shown, use z = z0+reit for t ∈ [θ1, θ2]
//
OO
◦z0
◦t = θ1
◦t = θ2
Some contours are naturally parametrised in more than one piece.
Example 11.6 To parametrise the unit square, with coordinates 0, 1, 1 + i, iwe can regard it as the sum of four line segments as shown.
//
OO
//
OOoo
��γ1
γ2
γ3
γ4
71
γ1: t for t ∈ [0, 1]γ2: 1 + it for t ∈ [0, 1]γ3: (1− t) + i for t ∈ [0, 1] (or t + i for t ∈ [1, 0])γ4: (1− t)i for t ∈ [0, 1] (or ti for t ∈ [1, 0])
One can run all these together into a single parametrisation:φ : [0, 4] −→ C defined by
φ(t) =
t if t ∈ [0, 1]1 + i(t− 1) if t ∈ [1, 2](1− (t− 2)) + i, = 3− t + i if t ∈ [2, 3]((1− (t− 3))i, = 4− t if t ∈ [3, 4]
but we will always find it more convenient to parametrise such pieces separately,so it will not be necessary to write down a single function to cover the wholecontour in such a case.
Example 11.7 The D-shaped contour shown naturally breaks into the straightline part γ1 which is t with t ∈ [−R,R] and the semicircle γ2 which is Reit witht ∈ [0, π].
//
OO
◦−R
◦R//
γ1
< γ2
12 Integration along parametrised contours
If C is a contour/curve in the complex plane and f is a function defined on Cwe may compute
∫C
f(z)dz as follows.Choose a parametrisation φ : [a, b] −→ C of C.Then ∫
C
f(z)dz =∫ b
a
f(φ(t))dφ(t)
dtdt,
that is, ∫ b
a
f(z)dz
dtdt
where z = φ(t).Of course, for this to make sense, f must be defined on C and the parametri-
sation must be a differentiable function (so that dφ(t)/dt makes sense) and onehas to accept that the answer is independent of the parametrisation chosen (itis).
72
Example 12.1 ∫C(0,1)
zndz
where n ∈ Z but n 6= −1.
Parametrise C(0, 1) as z = eit with t ∈ [0, 2π] so
dz
dt= ieit,
that isdz = izdt
(in the formula above we often replace dzdt dt by dz, which is computed exactly as
dzdt dt).
Then ∫=∫ 2π
0
(eit)nieitdt =∫ 2π
0
iei(n+1)tdt.
Here the variable t is real and, although complex numbers appear, they are con-stants, so it is valid to treat them just as we would treat real constants. So∫
=[
1n + 1
ei(n+1)t
]2π
0
=1
n + 1
(e2πi(n+1) − e0
)=
1n + 1
(1− 1) = 0
- which checks with our earlier direct computation of∫
C(0,1)z2dz.
Example 12.2 Now see what we get if we integrate zn (n 6= −1) just roundthe upper semi-circle eit, t ∈ [0, π].
The computation is just as before except with different limits of integration,and we obtain ∫
=[
1n + 1
ei(n+1)t
]π
0
=1
n + 1
(eπi(n+1) − ei0
)
=
1
n+1 (−1− 1) = −2n+1 if n is even,
1n+1 (1− 1) = 0 if n is odd.
Example 12.3 ∫l
Re(z)dz
where l is the line joining 0 to 1 + i.
Parametrise l as z = t(1 + i), t ∈ [0, 1] so dz = (1 + i)dt and Re(z) =Re(t + it) = t. Therefore∫
=∫ 1
0
t(1 + i)dt =[(1 + i)
t2
2
]10
=1 + i
2
73
Example 12.4 ∫S
zdz
where S is the square with corners 0, 1, 1 + i, i.
Break S into the four sides γ1, . . . , γ4 as before (Example 11.6).γ1: z = t, t ∈ [0, 1] so dt = dz and z = t = t, then∫
γ1
zdz =∫ 1
0
tdt =[t2
2
]10
=12
γ2: z = 1 + it, t ∈ [0, 1] so dz = idt and z = 1− it, therefore∫γ2
zdz =∫ 1
0
(1− it)idt =∫ 1
0
(i + t)dt =[it +
t2
2
]10
= i +12
γ3: z = t + i, t ∈ [1, 0] so dz = dt and z = t− i, so∫γ3
zdz =∫ 0
1
(t− i
)dt =
[t2
2− it
]01
= −12
+ i
γ4: z = ti, t ∈ [1, 0] so dz = idt and z = −ti, hence∫γ4
zdz =∫ 0
1
−tiidt =∫ 0
1
tdt =[t2
2
]01
= −12.
Therefore∫S
zdz =∫
γ1
zdz +∫
γ2
zdz +∫
γ3
zdz +∫
γ4
zdz =12
+ i +12− 1
2+ i− 1
2= 2i.
By way of contrast, you should, as an exercise, compute∫
Cz dz where C =
C(
12 + 1
2 i, 12
)is the circle inscribed in the square.
Exercise 12.5 Compute ∫γ
|z|2dz
where γ is the straight line from i to 1.
Exercise 12.6 Compute ∫S
z2dz
where S is the square as in Example 12.4
You have probably seen a result like the next one in the context of integrationof real functions.
74
Theorem 12.7 (Estimate Lemma) If f is a complex-valued function, if γ isa contour and if |f(z)| ≤ M for all z on γ then∣∣∣∣ ∫
γ
f(z)dz
∣∣∣∣ ≤ Mlγ
where lγ is the length of γ.
(We haven’t actually defined what we mean by the length of a curve in thecomplex plane - you can take it to mean the length of the curve regarded as acurve in the real plane.)
13 Theorems on integration
The theorems in this section allow many integrals to be determined with littleor no computation. They do, however, require some conditions on the functionbeing integrated and they do not help with integrating functions involving,for instance, |z|, Re(z), z, for which the methods of the previous section aregenerally needed.
Say that a complex function F is a primitive of the complex function f if
dF
dz= f :
for example zn+1/(n + 1) is a primitive of zn and ez is a primitive of ez (as isez + b for any constant b).
Theorem 13.1 If f has a primitive F in a region containing the contour γ(that is dF
dz = f(z) for all z in the region) and if γ is parametrised as φ(t) fort ∈ [a, b] then∫
γ
f(z)dz =∫ b
a
f(φ(t))dt =[F (φ(t))
]ba
= F (φ(b))− F (φ(a)).
Example 13.2 (cf. Example 12.2) Find∫γ
zndz
where n is an integer, n 6= −1 and where γ is the contour along the uppersemicircle with centre 0 and radius 1 going from 1 to −1
By Theorem 13.1∫=[
zn+1
n + 1
]−1
1
=1
n + 1((−1)n+1 − 1n+1)
75
=
0 if n is odd
− 2n+1 if n is even
Example 13.3 Integrate the same function as in Example 13.2 but round thesemicircle γ′ of radius 1 going anti-clockwise from i to −i.∫
γ′zndz =
[zn+1
n + 1
]−i
i
=1
n + 1((−i)n+1 − in+1
)
= 0 if n is odd (since then (−i)n+1 = (−1)n+1in+1 = in+1
= 2n+1 in+1 if n is even
{= − 2
n+1 i if n ≡ 0 mod 4= 2
n+1 i if n ≡ 2 mod 4
Example 13.4∫
γzn+1dz where γ is any contour joining i to −i: the answer
is exactly the same as in the previous example!
The example above illustrates the remarkable fact that if f has a primitive ina “simply-connected” region and γ is any contour within that region, with start-point za and end-point zb then the value of
∫γ
f(z)dz is F (zb) − F (za), henceis independent of the path taken in going from a to b (so long as it stays in theregion). This fact is highly relevant to many situations in physics. By “simplyconnected” we mean, roughly, that the region does not enclose any “holes”. Forinstance, the complex plane with zero removed is not simply connected and thisis why integrating a function like 1/z round the unit circle (which contains thehole) does not give the answer zero (whereas integrating 1/z round a contourwhich does not contain its singularity at 0 does give zero). Even if the regionis not simply-connected the path-independence of the value of the integral isstill true for paths which are “homotopic”, meaning that the one path can becontinuously deformed to the other while staying within the region.
Applied to a closed contour 13.1 gives the following corollary.
Theorem 13.5 If f has a primitive in a region containing the closed contourγ then ∫
γ
f = 0.
Proof. Regard γ as a curve with start- and end-point being za (any point onthe contour). By Theorem 13.1
∫γ
f = F (za) − F (za) = 0, where F is anyprimitive of f . 2
Example 13.6 The fact that∫
C(0,1)zndz = 0 for n ∈ Z, n 6= −1 which was
shown earlier (Example 12.1), is a direct consequence.
76
The situation for n = −1 is different because the function z−1 does not havea primitive on the unit circle: recall that log(z) is many-valued and running onceround the unit circle, trying to choose a value of log in a continuous way, meansthat the value of log, once we have returned to the start, has increased by 2πi.Thinking in terms of the Riemann surface which is the graph of the exponentialfunction, one might guess that, since this is the increase in the value of logobtained by running once round the circle, this will be the value of
∫C(0,1)
1z dz
and this is, indeed the case, as we check now.
Example 13.7 ∫C(0,1)
1zdz
Method 1: parametrise C(0, 1) as z = eit, t ∈ [0, 2π] so dz = ieitdt and∫=∫ 2π
0
e−itieitdt =∫ 2π
0
idt = 2πi.
Method 2: split C(0, 1) into the upper, γu, and lower, γl, semicircles. On eachof these contours the function 1/z does have a primitive:we may choose the primitive on γu to be the principal value of log, namelylog|z|+ iarg(z) where arg(z) ∈ [0, π];and on γl we may choose to use log|z|+ iarg(z) + 2πi where arg(z) ∈ [−π, 0] asprimitive.Note that these choices match up at z = −1 so that we have chosen a continuousprimitive which is defined on the Riemann surface rather than on the complexplane.
Then ∫C(0,1)
1zdz =
∫γu
1z
+∫
γl
1zdz
= {log| − 1|+ iarg(−1)− (log|1|+ iarg(1))}+ {(log|1|+ iarg(1) + 2πi)− (log| −1|+ iarg(−1) + 2πi)}
= {log(1) + πi− log(1)− 0}+ {log(1) + 0 + 2πi− log(1)− i(−π)− 2πi}= πi + πi = 2πi
For functions which do not have primitives on a contour, or where the prim-itive might be difficult to compute explicitly (that is, where it might be difficultto actually “do the integration”) the next theorem often applies.
Theorem 13.8 (Cauchy’s Theorem) Suppose γ is a closed contour containedin a region and suppose that the complex function f is differentiable on andinside γ. Then ∫
γ
f(z)dz = 0.
77
Example 13.9∫
C(0,1)zndz yet again.
Cauchy’s Theorem does not apply if n < 0 (since then zn has a singularityat 0 so certainly is not differentiable at 0). But if n is an integer with n ≥ 0then zn is differentiable on and inside the unit circle so by Cauchy’s Theorem,∫
C(0,1)
zndz = 0.
(The Residue Theorem, 13.13, below is a result that will apply to the negativepowers as well.)
Corollary 13.10 If γ1, γ2 are paths with the same start and end points and ifeither:
(i) f has a primitive in a region containing both γ1 and γ2; or(ii) f is differentiable in a region containing γ1 and γ2 and every point
between them;then ∫
γ1
fdz =∫
γ2
fdz.
Proof.
◦
◦>γ1
>γ2
Consider the closed path γ1−γ2 and apply 13.5 or Cauchy’s Theorem (13.8),as appropriate. 2
By a simple closed contour we will mean a contour which is closed butdoes not otherwise intersect itself. This excludes, for instance, a circle runround more than once as well as self-intersecting curves. This is to simplify thestatements of the remaining results of this section: otherwise we would have tointroduce “winding numbers” of curves round points.
Corollary 13.11 If γ1, γ2 are simple closed contours, if γ1 lies inside γ2 andhas the same orientation as γ2, and if f is differentiable in a region containingγ1 and γ2 and all the points between them then
∫γ1
fdz =∫
γ2fdz.
Proof. Connect γ1 and γ2 by paths γ3 and γ4 and break up γ1 as γ′1 + γ′′1 andγ2 as γ′2 + γ′′2 (as shown).
78
_ekr{����� % + 3:DLSY _ ekrz�����%+2:DLSY
_behknruy��������� " % ( +.25:?DILPSVY\ _ behknruz���������"%(+.25:?DILPSVY\
___ > ___ >
>
<
>
<
γ3 γ4
γ′′2
γ′′1
γ′2
γ′1
Then we have, by 13.10,∫
γ′1= −
∫γ4
+∫
γ′2−∫
γ3and
∫γ′′1
=∫
γ3+∫
γ′′2+∫
γ4.
Adding these together we obtain∫
γ1=∫
γ′1+∫
γ′′1=∫
γ′2+∫
γ′′2=∫
γ2. 2
This method of proof extends to give the following.
Corollary 13.12 Suppose that γ1, . . . , γn are simple closed contours which donot intersect and none of which is contained in any other. Suppose that all ofγ1, . . . , γn are contained in the simple closed contour γ and all are given theorientation induced by γ. Let f be differentiable in a region containing all thesecurves and all the points in between γ and the others. Then∫
γ
fdz =n∑
k=1
∫γk
fdz.
Theorem 13.13 (Residue Theorem) Suppose that γ is a simple closed contourand f is a function which is differentiable on and inside γ except possibly atfinitely many points c1, . . . cm, where it has poles. Then∫
γ
fdz = 2πi
m∑k=1
res(f, ck).
Proof. First we consider the case that f has just one singularity, z = c, insideγ. Choose r small enough that the circle, γ′ = C(c, r), with centre c and radiusr lies entirely inside γ. Then by 13.11
∫γ
fdz =∫
γ′fdz.
Suppose that the Laurent expansion of f(z) around c is
f(z) =∞∑−∞
an(z − c)n.
Then ∫γ′
fdz =∫
γ′
∞∑−∞
an(z − c)n
79
=∞∑−∞
an
∫γ′
(z − c)n = a−1
∫γ′
dz
z − c
because, as we have seen,∫
γ′(z − c)ndz = 0 for n 6= −1 (and because it is the
case that complex-differentiable functions can be integrated by integrating theirpower series expansions term-by-term).
Parametrise γ′ as z = c + reit (0 ≤ t ≤ 2π) and then∫γ′
dz
z − c=∫ 2π
0
ireitdt
reit=∫ 2π
0
idt = 2πi.
Therefore∫
γfdz =
∫γ′
fdz = a−12πi, as required.
For the general case, with more than one singularity, put a little circle γk
round each of the poles ck, choosing these small enough so that none intersectand none is inside any other.
_abdfgikmoqsvy|����������� "$')+.137:=ADGILNPRTVXY[ ] ^ ` a cefhjlnpruwz}���� �������!#%(*-/258;?CFHKMOQSUWXZ\]
◦c1
◦c2
◦c3
γ1
γ2
γ3
Then use 13.12 to conclude∫γ
fdz =m∑
k=1
2πires(f, ck) = 2πim∑
k=1
res(f, ck).
2
Example 13.14∫
C(0,1)z−ndz for n ∈ N
The function z−n has just one pole, at 0, inside C(0, 1) and is differentiableon and in C(0, 1) at every other point. To find res(z−n, 0) we need the Laurentexpansion of z−n about 0: but that is already in the form of a Laurent expansion,so
res(z−n, 0) ={
1 if n = 1,= 0 otherwise
and hence, by the Residue Theorem,∫C(0,1)
z−n ={
2πi if n = 1,= 0 otherwise
80
Example 13.15 ∫γ
dz
z2 + 1=∫
γ
dz
(z − i)(z + i)
This function has two poles, one at i, where the residue is 1/(2i), the otherat −i, where the residue is −1/(2i).
There are only four possible answers, therefore, for∫
γdz/(z2 +1), depending
on which of these poles, if any, are inside γ.
Case 1: neither i nor −i is inside γ.By Cauchy’s Theorem ∫
γ
dz
z2 + 1= 0.
//
OO
◦i
◦−i
_bdgjmqv{����" (.4;CIMQTWZ\ _ bdgjmpu{����"(.4;BHMQTWZ\
Case 2: i but not −i is inside γ.By the Residue Theorem∫
γ
dz
z2 + 1= 2πi · 1
2i= π.
//
OO
◦i
◦−i
_bdgjmqv{����" (.4;CIMQTWZ\ _ bdgjmpu{����"(.4;BHMQTWZ\
Case 3: −i but not i is inside γ.By the Residue Theorem∫
γ
dz
z2 + 1= 2πi · −1
2i= −π.
81
//
OO
◦i
◦−i
_fnw����� $ ( /4>GPX_fow�����$(.4>GPX
Case 4: both i and −i are inside γ.By the Residue Theorem∫
γ
dz
z2 + 1= 2πi
( 12i
+−12i
)= 0.
//
OO
◦i
◦−i
_cfjnrw}�� ����� ! % ( + /48>DIMRVZ]adhlpuz������� #&*-16;AFKPTX[_
(Note that if either i or −i is on γ then∫
γdz/(z2 + 1) is not defined since
the function is undefined at a point actually on the contour.)
We give one notable consequence of the residue theorem.
Theorem 13.16 Cauchy’s Integral Formula) Let γ be a simple closed contour.Suppose that f is complex-differentiable in a region which contains γ and theinterior of γ. Let a be a point in the interior of γ. Then
12πi
∫γ
f(z) dz
z − a= f(a).
Proof. The function f(z)/(z − a) has a simple pole at z = a and there itsresidue is f(a) so, by the Residue Theorem
∫γ
f(z)dz/(z − a) = 2πif(a), whichgives the result. 2
Note the following interpretation of this result. The value at z = a of acomplex- differentiable function f can be computed by averaging f(z)/(z − a)(=the value of f at z, divided by the difference, z − a) on any simple closedcontour containing a and then dividing by 2πi. In particular, the value of acomplex-differentiable function at any point can be found from the values onany surrounding contour (possibly distant from a).
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14 Real trigonometric integrals via complex num-bers
The general idea here is to convert a real integral over a finite interval into acomplex integral over the unit circle which is easier to evaluate. We illustratewith one example. More examples will be given in lectures and there are morein Reade’s book.
Example 14.1 ∫ 2π
0
dt
5 + 4cos(t)
We turn the real interval [0, 2π] into the unit circle in the complex plane bythe substitution z = eit (as t varies from 0 to 2π, z runs anticlockwise over theunit circle: that is, this is a parametrisation of the unit circle).
Since z = eit, dz/dt = ieit = iz, so we can write dz = izdt, that is dz/(iz) =dt.
We have to write everything in terms of the new variable z, in particularwe still have to write cos(t) in terms of z. But recall Euler’s formula: cos(t) =12 (eit − e−it), that is cos(t) = 1
2 (z + z−1).So ∫
=∫
C(0,1)
dz/(iz)5 + 4 · 1
2 (z + z−1)=∫
C(0,1)
dz
iz(5 + 2z + 2z−1)
where C(0, 1) denotes the unit circle (with anticlockwise orientation)
=1i
∫C(0,1)
dz
2z2 + 5z + 2=
1i
∫C(0,1)
dz
(2z + 1)(z + 2)
The last integral is of a sort that we have met before and to which we can applythe Residue Theorem.
The singularities of the function f(z) = 1(2z+1)(z+2) are at z = −1/2 and
z = −2. Remember that only singularities inside the contour, C(0, 1), con-tribute to the integral, so we can ignore the singularity at z = −2. We haveRes(f,−1/2) = 1/3 (using either of the “quick methods”: note that to use thefirst method we must write f(z) as
1(z + 1
2 )2(z + 2)).
Therefore ∫=
1i2πi
13
=2π
3.
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15 Infinite real integrals
We illustrate with one example. Again, others (including variations) will begiven in lectures and can be found in Reade’s book.
Example 15.1 ∫ ∞
−∞
dx
x2 + 1
We consider a D-shaped contour, γ, of the complex plane as shown.
//
OO
◦−R
◦R//
γ1
< γ2
So γ = γ1+γ2 where γ1 is the (real) interval [−R,R] and γ2 is the semicirclein the upper half-plane with centre 0 and radius R. Here R is a positive realnumber which will go off to infinity in the later part of the analysis, so think ofR as “large”.
Note that ∫γ
dz
z2 + 1=∫
γ1
dz
z2 + 1+∫
γ2
dz
z2 + 1(∗)
On the contour γ1 we have∫γ1
dz
z2 + 1=∫ R
−R
dx
x2 + 1
which, if we let R →∞, will give the value of∫ ∞
−∞
dx
x2 + 1
if this limit exists. (Well, there is the point that the right- and left-hand limitsshould go to infinity separately, consider
∫∞−∞ xdx, but, for this (everywhere
positive) function, it is valid to compute it this way.)The idea is to separately compute the other two terms,
∫γ2
and∫
γ, in equation
(*) and then let R →∞.In fact, we estimate, rather than compute,
∫γ2
: so it will be necessary thatthe value of this integral tend to a definite limit (which will be 0) in order forthis method to work. We use the Estimate Lemma 12.7. The length of the
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contour γ2 is just πR. Also, for z on the contour γ2 we have |z| = R so|z2 + 1| ≥ |z2| − 1 = R2 − 1 and hence∣∣∣∣ 1
z2 + 1
∣∣∣∣ ≤ 1R2 − 1
(assume R is large enough so that R2 − 1 > 0). So the Estimate Lemma gives∣∣∣∣ ∫γ2
dz
z2 + 1
∣∣∣∣ ≤ πR
R2 − 1.
As R →∞ this tends to 0 (so that’s one term taken care of).As for
∫γ, we can use the Residue Theorem 13.13. The singularities of
1/(z2 + 1) = 1/((z + i)(z − i)) occur at i and −i. Only singularities insidethe contour are relevant so forget about −i. We’re going to let R → ∞ somay as well suppose that R is big enough for i to be inside the contour. NowRes(1/(z2 + 1), i) = 1/(2π) so by the Residue Theorem,∫
γ
dz
z2 + 1= 2πi
12i
= π.
Therefore, for R large enough, equation (*) becomes
π =∫
γ1
dz
z2 + 1+∫
γ2
dz
z2 + 1,
that is ∫γ1
dz
z2 + 1= π −
∫γ2
dz
z2 + 1.
Now let R →∞. Then
π −∫
γ2
dz
z2 + 1→ π − 0 = π
and ∫γ1
dz
z2 + 1→∫ ∞
−∞
dx
x2 + 1.
We deduce that ∫ ∞
−∞
dx
x2 + 1= π.
We will see, in the lectures, variations of this. For example, if the functionto be integrated has a singularity on the real axis then we can try putting alittle loop around it. In that case we will find the following theorem of use.
Theorem 15.2 (Half-residue Theorem) If γr is the contour z = c + reit (t ∈[0, π]) (upper semi-circle, centre c ∈ C, radius r) and if f has, near c, only asimple pole, at c, then∫
γr
f(z)dz → πi res(f, c) as r → 0.
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